Normal distribution:
A normal distribution is a continuous probability distribution of a random variable whose graph is a symmetric bell-shaped curve. It is completely determined by two parameters: mean $\mu$ and standard deviation $\sigma$. Its probability density function is

$f(x)=\frac{1}{\sigma\sqrt{2\pi}}e^{-\frac{(x-\mu)^2}{2\sigma^2}},\qquad -\infty<x<\infty.$

Properties: mean, median and mode are equal; the curve is symmetric about the mean; total area under the curve is 1; about 68% of values lie within $\mu\pm\sigma$, about 95% within $\mu\pm2\sigma$, and about 99.7% within $\mu\pm3\sigma$.

Given: $\mu=72,\qquad \sigma=7.5$

Standard normal variable: $Z=\frac{X-\mu}{\sigma}=\frac{X-72}{7.5}$

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(i) Percentage of students in grade A

Grade A means score $X\ge 80$.

$P(X\ge 80)=P\left(Z\ge \frac{80-72}{7.5}\right)$ $=P(Z\ge 1.067)$ $=1-\Phi(1.067)$ $\approx 1-0.8569=0.1431$

So, $0.1431\times 100\%=14.31\%$

Therefore, about 14.31% of students get grade A.

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(ii) Lowest score of the top 10% students

Top 10% students means the 90th percentile. Let the lowest score be $x$.

$P(X\ge x)=0.10$ $P(X\le x)=0.90$

For standard normal distribution, $P(Z\le z_{0.90})=0.90$ $z_{0.90}\approx 1.2816$

Now, $x=\mu+z\sigma$ $x=72+1.2816(7.5)$ $x=72+9.612$ $x=81.612$

Therefore, the lowest score of the top 10% students is approximately 81.61 marks.
If marks are taken as whole numbers, the lowest integer score is 82 marks.

The population standard deviation is not given, so the sample standard deviation is used and a one-sample $t$-test is applied (the population is normal, $n=10$).

(i) Test of hypothesis

$H_0:\mu=100$ against $H_1:\mu\neq100$ (two-tailed), $\alpha=0.05$.

$\bar x=\frac{985}{10}=98.5$

$\sum(x-\bar x)^2 = 6238.5$, so

$s^2=\frac{6238.5}{9}=693.17,\qquad s=26.33$

$t=\frac{\bar x-\mu_0}{s/\sqrt n}=\frac{98.5-100}{26.33/\sqrt{10}}=\frac{-1.5}{8.33}\approx-0.18$

Critical value: $t_{0.025,\,9}=2.262$. Since $|t|=0.18<2.262$, $H_0$ is not rejected. There is no significant evidence at the 5% level that sufferers differ from the general population in their test performance.

(ii) 95% confidence limits for the true mean

$\bar x\pm t_{0.025,9}\frac{s}{\sqrt n}=98.5\pm2.262\times8.33=98.5\pm18.8$

The 95% confidence limits are approximately $(79.7,\ 117.3)$. The value $100$ lies within this interval, which agrees with the test.

Let
[ x=\text{Dosage},\qquad y=\text{Duration of relief} ]

Given: [ n=8,\quad \sum x=56,\quad \sum y=140,\quad \sum x^2=424,\quad \sum y^2=2676,\quad \sum xy=1053 ]

Also, [ \bar x=\frac{56}{8}=7,\qquad \bar y=\frac{140}{8}=17.5 ]

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i. Relationship between (x) and (y)

The relationship is measured by Karl Pearson’s correlation coefficient: [ r=\frac{n\sum xy-\sum x\sum y} {\sqrt{\left[n\sum x^2-(\sum x)^2\right]\left[n\sum y^2-(\sum y)^2\right]}} ]

Substituting: [ r=\frac{8(1053)-(56)(140)} {\sqrt{\left[8(424)-56^2\right]\left[8(2676)-140^2\right]}} ]

[ r=\frac{8424-7840}{\sqrt{(3392-3136)(21408-19600)}} ]

[ r=\frac{584}{\sqrt{256\times 1808}} =\frac{584}{680.33} ]

[ \boxed{r=0.8584} ]

Interpretation

There is a strong positive correlation between dosage and duration of relief. As dosage increases, the duration of relief also tends to increase. Since (r^2=0.7369), about 73.69% of the variation in duration of relief is explained by dosage.

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ii. Regression equation of (y) on (x)

The regression line is: [ y=a+bx ]

where [ b=\frac{n\sum xy-\sum x\sum y}{n\sum x^2-(\sum x)^2} ]

[ b=\frac{584}{256}=2.28125 ]

And [ a=\bar y-b\bar x ]

[ a=17.5-(2.28125)(7) ]

[ a=17.5-15.96875=1.53125 ]

Hence the regression equation is: [ \boxed{y=1.53125+2.28125x} ]

or approximately, [ \boxed{y=1.53+2.28x} ]

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iii. Estimate (y) when (x=8)

Using [ y=1.53125+2.28125x ]

[ y=1.53125+2.28125(8) ]

[ y=1.53125+18.25 ]

[ \boxed{y=19.78125\approx 19.78} ]

So when dosage is (8), the estimated duration of relief is about 19.78 units.

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iv. Interpretation of the slope

The slope is [ b=2.28125 ]

It means that for every one unit increase in dosage, the duration of relief increases on average by about 2.28 units.

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v. Standard error of estimate

The standard error of estimate is: [ S_e=\sqrt{\frac{\sum y^2-a\sum y-b\sum xy}{n-2}} ]

Using the regression values: [ S_e=\sqrt{\frac{226-(2.28125)(73)}{6}} ]

Since [ \sum y^2-\frac{(\sum y)^2}{n}=2676-\frac{140^2}{8}=226 ]

and [ b\sum xy=2.28125(73)=166.53125 ]

[ S_e=\sqrt{\frac{226-166.53125}{6}} ]

[ S_e=\sqrt{\frac{59.46875}{6}} ]

[ S_e=\sqrt{9.911458} ]

[ \boxed{S_e=3.148} ]

Therefore, the standard error of estimate is approximately 3.15 units.

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Final Answers

[ \boxed{r=0.8584} ]

Strong positive correlation between dosage and duration of relief.

[ \boxed{y=1.53125+2.28125x} ]

When (x=8), [ \boxed{y\approx 19.78} ]

Slope interpretation: For each additional unit of dosage, duration of relief increases by about 2.28 units.

[ \boxed{S_e\approx 3.15} ]

Let (X) be the number of defective parts among the two randomly selected parts.

Given: [ n=2,\qquad p=P(\text{defective})=0.03,\qquad q=P(\text{non-defective})=0.97 ]

i. Conditions for a binomial experiment

The situation is a binomial experiment if:

1. Fixed number of trials: Exactly (n=2) parts are selected.
2. Two possible outcomes per trial: Each part is either defective or non-defective.
3. Independent trials: The condition of one part does not affect the condition of the other part. This is reasonable if the machine is functioning properly and the production process is stable.
4. Constant probability of success: The probability that a part is defective remains the same for each part, (p=0.03).
5. Random variable counts successes: (X) counts the number of defective parts found.

Thus, [ X \sim \text{Binomial}(n=2,\ p=0.03) ]

ii. Probabilities

The binomial probability formula is [ P(X=x)={}^{n}C_x p^x q^{n-x} ]

No defective part: (X=0)

[ P(X=0)={}^{2}C_0(0.03)^0(0.97)^2 ] [ =1\times 1\times 0.9409=0.9409 ] [ \boxed{P(X=0)=0.9409 \approx 94.09%} ]

One defective part: (X=1)

[ P(X=1)={}^{2}C_1(0.03)^1(0.97)^1 ] [ =2\times 0.03\times 0.97=0.0582 ] [ \boxed{P(X=1)=0.0582 \approx 5.82%} ]

Two defective parts: (X=2)

[ P(X=2)={}^{2}C_2(0.03)^2(0.97)^0 ] [ =1\times 0.0009\times 1=0.0009 ] [ \boxed{P(X=2)=0.0009 \approx 0.09%} ]

Probability distribution table

So the probabilities are: [ \boxed{P(0)=0.9409,\quad P(1)=0.0582,\quad P(2)=0.0009} ]

Interval estimation: Interval estimation is the method of estimating an unknown population parameter by a range of values, called a confidence interval, that is expected to contain the parameter with a stated degree of confidence (e.g. 95%). It gives a lower and an upper confidence limit instead of a single point estimate.

Note on the data: the question as recorded gives $\sigma = 50$ hours but not the sample size $n$ or the sample mean $\bar x$. The general formulas are given first, and a numerical illustration follows. Substitute your own $n$ and $\bar x$.

Formulas (population standard deviation known, 95% confidence, $z_{0.025} = 1.96$):

$\text{(i)}\ SE = \frac{\sigma}{\sqrt n} = \frac{50}{\sqrt n}$

$\text{(ii)}\ E = z_{\alpha/2}\times SE = 1.96 \times \frac{50}{\sqrt n}$

$\text{(iii)}\ \text{CI} = \bar x \pm E$

Illustration (taking $n = 100$ and $\bar x = 780$ hours):

$SE = \frac{50}{\sqrt{100}} = 5 \text{ hours}, \qquad E = 1.96 \times 5 = 9.8 \text{ hours}$

$\text{CI} = 780 \pm 9.8 = (770.2,\ 789.8) \text{ hours}$

We are 95% confident that the mean life of all bulbs produced by the firm lies between 770.2 and 789.8 hours.

Given:
Mean, $\mu = 10$ cm
Standard deviation, $\sigma = 0.04$ cm

Let $X$ be the inside diameter of a piston ring.
$X \sim N(10,\ 0.04^2)$

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(i) Proportion of rings with inside diameter exceeding 10.08 cm

We need: $P(X > 10.08)$

Standardize: $z = \frac{X - \mu}{\sigma} = \frac{10.08 - 10}{0.04} = \frac{0.08}{0.04} = 2$

So, $P(X > 10.08) = P(Z > 2)$

From the standard normal table: $P(Z > 2) = 1 - P(Z \le 2) = 1 - 0.9772 = 0.0228$

Therefore, the proportion of rings exceeding 10.08 cm is:

$\boxed{0.0228 \text{ or } 2.28\%}$

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(ii) Value below which 10% of the piston rings fall

We need $x$ such that: $P(X < x) = 0.10$

Standardize: $P\left(Z < \frac{x - 10}{0.04}\right) = 0.10$

Let: $z = \frac{x - 10}{0.04}$

From the standard normal table, the value of $z$ corresponding to a left-tail area of $0.10$ is: $z = -1.28$

So: $\frac{x - 10}{0.04} = -1.28$

$x - 10 = -1.28 \times 0.04$

$x - 10 = -0.0512$

$x = 10 - 0.0512 = 9.9488$

Therefore, 10% of the piston rings will have an inside diameter below:

$\boxed{9.9488 \text{ cm} \approx 9.95 \text{ cm}}$

(i) Association of Attributes

When two or more attributes (qualitative characteristics such as literacy, sex, employment) are observed on the same set of units, we may ask whether they are associated — i.e. whether the presence of one attribute influences the presence of the other. If the distribution of one attribute is the same in every class of the other, the attributes are independent; otherwise they are associated (positively or negatively).

Association is studied by arranging the data in a 2 × 2 contingency table:

where A, B denote the presence and α, β the absence of the attributes.

Methods of studying association:

• Proportion method — compare the proportion of B among A with the proportion of B among α.
• Yule's coefficient of association: $Q = \frac{ad - bc}{ad + bc}$ $Q$ ranges from $-1$ (perfect negative association) through $0$ (independence) to $+1$ (perfect positive association).
• Yule's coefficient of colligation: $Y = \frac{\sqrt{ad} - \sqrt{bc}}{\sqrt{ad} + \sqrt{bc}}$
• Chi-square test for independence in larger tables.

Association of attributes is widely used in sociology, education and medical statistics — for example, to test whether literacy and employment status are related.

(ii) Nominal Scale and Ordinal Scale

Nominal scale is the weakest level of measurement. It merely names or labels the categories to which objects belong; the numbers or labels have no numerical meaning and cannot be ordered, added or averaged. Categories are mutually exclusive and collectively exhaustive. Examples: sex (male/female), religion, blood group, marital status, district of residence. The only permissible operations are counting frequencies and finding the mode; statistical tools used are proportions, percentages and the chi-square test.

Ordinal scale (also called ranking scale) possesses the property of order in addition to naming. The categories can be arranged in a meaningful rank order, but the differences between ranks are not equal or measurable. Examples: socio-economic status (low/middle/high), Likert-scale responses (strongly disagree … strongly agree), examination grades, severity of a disease. Besides counting, we can compute the median, quartiles and rank-correlation coefficients; arithmetic mean is not strictly valid.

Key difference: nominal scale only classifies, while ordinal scale classifies and orders. Neither permits the arithmetic operations allowed by interval and ratio scales.

(iii) Type I Error and Type II Error in Testing of Hypothesis

In hypothesis testing we set up a null hypothesis $H_0$ and an alternative $H_1$, and decide to accept or reject $H_0$ on the basis of sample evidence. Since the decision is based on a sample, two kinds of errors are possible:

Type I error: rejecting $H_0$ when it is actually true — i.e. rejecting a true null hypothesis. Its probability is denoted by $\alpha$, the level of significance (commonly 0.05 or 0.01). It is also called the producer's risk or error of the first kind. Example: concluding that a medicine is effective when it is actually not.

Type II error: accepting $H_0$ when it is actually false — i.e. failing to reject a false null hypothesis. Its probability is denoted by $\beta$. The quantity $1 - \beta$ is the power of the test, the probability of correctly rejecting a false $H_0$. It is also called the consumer's risk or error of the second kind. Example: concluding that a medicine is ineffective when it actually works.

Relation between the two: for a fixed sample size, decreasing $\alpha$ (making rejection harder) increases $\beta$, and vice versa; the two errors cannot both be minimised simultaneously. The usual practice is to fix $\alpha$ at a small value and then choose the test that minimises $\beta$, or to increase the sample size so that both errors are reduced.

Assumptions of (simple linear) regression analysis

1. Linearity: the relationship between $X$ and the mean of $Y$ is linear, $Y = \alpha + \beta X + \varepsilon$.
2. Independence: the errors $\varepsilon_i$ are independent of each other (no autocorrelation).
3. Homoscedasticity: the errors have constant variance $\sigma^2$ for all values of $X$.
4. Normality: the errors are normally distributed with mean zero, $\varepsilon \sim N(0, \sigma^2)$.
5. $X$ is measured without error (fixed or non-random), and there is no perfect multicollinearity (for multiple regression).

Calculation

Let $X$ = mileage (1000 km) and $Y$ = sale price (100 $);$n = 5$.

$\bar X = 340/5 = 68$ and $\bar Y = 340/5 = 68$.

i. Relationship (correlation):

$r = \frac{\sum xy}{\sqrt{\sum x^2 \sum y^2}} = \frac{-290}{\sqrt{716\times130}} = \frac{-290}{305.1} = -0.95$

There is a strong negative linear relationship: the higher the mileage, the lower the sale price. About $r^2 = 90\%$ of the variation in price is explained by mileage.

ii. Least-squares regression line $\hat Y = a + bX$:

$b = \frac{\sum xy}{\sum x^2} = \frac{-290}{716} = -0.405$

$a = \bar Y - b\bar X = 68 - (-0.405)(68) = 95.54$

$\boxed{\hat Y = 95.54 - 0.405X}$

iii. Interpretation of the slope: for each additional 1000 km driven, the sale price falls by about 0.405 hundred dollars, i.e. about USD 40.5, on average.

iv. Prediction for 70,000 miles: the data are in 1000 km, so convert first: $70{,}000\ \text{miles} \times 1.609 = 112{,}654\ \text{km} \approx 112.65$ thousand km.

$\hat Y = 95.54 - 0.405(112.65) \approx 49.9$

Predicted price $\approx 49.9 \times 100 \approx$ USD 4,990.

This lies outside the observed range of mileage (55 to 89 thousand km), so it is an extrapolation and should be treated with caution. (If 70 were read directly as 70 thousand km, the prediction would be $95.54 - 0.405\times 70 = 67.2$, i.e. about USD 6,720.)

Chi-square test of independence

Given data

Hypotheses

$H_0$: There is no difference among the age groups in the proportion of social media users who use a mobile phone to access social networking (the two variables are independent).

$H_1$: There is a difference among the age groups in the proportion of social media users who use a mobile phone to access social networking (the two variables are not independent).

Test statistic

$\chi^2 = \sum \frac{(O_{ij} - E_{ij})^2}{E_{ij}}, \qquad E_{ij} = \frac{R_i \times C_j}{n}$

Degrees of freedom $= (r-1)(c-1) = (2-1)(3-1) = 2$

Expected frequencies

$E_{11} = \dfrac{150 \times 100}{300} = 50$, and similarly every expected frequency is 50:

Computation of $\chi^2$

Critical value

At $\alpha = 0.05$ and $df = 2$, $\chi^2_{0.05,\,2} = 5.991$.

Decision

Since $\chi^2_{cal} = 78.08 > \chi^2_{crit} = 5.991$, we reject $H_0$ at the 5% level of significance.

Conclusion

There is sufficient evidence at the 0.05 level of significance of a difference among the age groups with respect to the use of a mobile phone for accessing social networking. The 65+ group uses mobile phones far less (only 14%) than the 18–34 (70%) and 35–64 (66%) groups.

Note on the data: the reaction-time table is missing from the question as recorded. The working below uses the usual textbook data for five subjects (before: 20, 18, 25, 22, 15; after: 22, 19, 26, 25, 18) as an illustration; the method is identical for any table.

1. Hypotheses

Let $d = X_{\text{after}} - X_{\text{before}}$, so a positive $d$ means the drug increased reaction time.

$H_0: \mu_d = 0 \quad \text{(no effect)} \qquad H_1: \mu_d > 0 \quad \text{(the drug increases reaction time)}$

This is a right-tailed paired $t$-test at $\alpha = 0.05$.

2. Data

3. Test statistic

$\bar d = \frac{10}{5} = 2, \qquad s_d^2 = \frac{24 - 10^2/5}{4} = 1, \qquad s_d = 1$

$t = \frac{\bar d}{s_d/\sqrt n} = \frac{2}{1/\sqrt5} = 2\sqrt5 = 4.472$

4. Decision

Degrees of freedom $= 4$, and the right-tailed critical value is $t_{0.05,4} = 2.132$. Since $4.472 > 2.132$, we reject $H_0$.

5. Conclusion

At the 5% level there is sufficient evidence that the drug increases reaction time.

Conditional Probability

The conditional probability of an event $A$ given that an event $B$ has already occurred is defined as

$P(A \mid B) = \frac{P(A \cap B)}{P(B)}, \quad \text{provided } P(B) > 0.$

It measures the probability of $A$ when the sample space is restricted to the event $B$.

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Numerical

Given:

• $P(M) = 0.60$ (male), $P(F) = 0.40$ (female)
• $P(A \mid M) = 0.25$ (male getting grade A)
• $P(A \mid F) = 0.20$ (female getting grade A)

To find: $P(F \mid A)$

Step 1: Probability that a randomly selected student got grade A

$$\begin{aligned} P(A) &= P(A \mid M)\,P(M) + P(A \mid F)\,P(F) \\ &= (0.25)(0.60) + (0.20)(0.40) \\ &= 0.15 + 0.08 = 0.23 \end{aligned}$$

Step 2: By Bayes' theorem

$$P(F \mid A) = \frac{P(A \mid F)\,P(F)}{P(A)} = \frac{(0.20)(0.40)}{0.23} = \frac{0.08}{0.23} = 0.3478$$

Result:

$P(F \mid A) \approx 0.348 \approx 34.8\%$

Hence, given that the selected student got an "A" grade, the probability that the student is female is approximately 0.348 (34.8%).

Given: $n = 500$, $x = 340$.

i. Point estimate of the population proportion:

$\hat p = \frac{340}{500} = 0.68$

ii. Margin of error at 90% confidence: $z_{0.05} = 1.645$ and $\hat q = 0.32$.

$E = z_{\alpha/2}\sqrt{\frac{\hat p\hat q}{n}} = 1.645\sqrt{\frac{0.68\times0.32}{500}} = 1.645\sqrt{0.0004352} = 1.645\times0.02086 \approx 0.0343$

iii. 90% confidence interval:

$\hat p \pm E = 0.68 \pm 0.0343 = (0.6457,\ 0.7143)$

We are 90% confident that between about 64.6% and 71.4% of adults regard the future health of social security as a major economic concern.

Hypotheses:

$H_0$: the theoretical model fits the observed data well. $H_1$: the model does not fit the data well.

Calculation of $\chi^2 = \sum \dfrac{(O-E)^2}{E}$:

$\chi^2_{cal} \approx 10.137$

Degrees of freedom: 9 classes and no parameters estimated from the data, so $\nu = 9 - 1 = 8$.

Critical value: $\chi^2_{0.05,\,8} = 15.507$.

Decision: Since $10.137 < 15.507$, we fail to reject $H_0$.

Conclusion: At the 5% level we cannot conclude that the theoretical model fails to explain the observed values; the model fits the data adequately.

Short notes on any two:

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(i) Nominal Scale and Ordinal Scale

Nominal Scale: The nominal scale is the simplest and most elementary level of measurement. It is used merely to classify or identify objects, individuals or responses into mutually exclusive and collectively exhaustive categories, without any order or ranking. Numbers or labels are assigned only as names or tags. For example, gender (male/female), religion (Hindu/Muslim/Christian), blood group (A/B/AB/O) and marital status. The only mathematical operations possible are counting the frequency in each category; arithmetic operations like addition or subtraction are meaningless. Statistical measures such as mode and chi-square test are applicable.

Ordinal Scale: The ordinal scale is a scale of measurement in which the categories possess a meaningful order or rank, but the differences between the ranks are not equal or quantitatively defined. For example, ranks in an examination (first, second, third), socio-economic status (low, middle, high), and customer satisfaction (poor, fair, good, excellent). Here we can say that one category is greater or smaller than another, but we cannot say by how much. Besides the mode, the median and other order-based measures (such as rank correlation, Spearman's coefficient) are applicable. Thus, the ordinal scale is a higher level of measurement than the nominal scale, as it possesses the property of order in addition to classification.

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(ii) Poisson Distribution

The Poisson distribution is a discrete probability distribution that expresses the probability of a given number of events occurring in a fixed interval of time or space, when these events occur with a known average rate and independently of the time since the last event. It is named after the French mathematician S. D. Poisson.

Definition: A random variable $X$ is said to follow a Poisson distribution with parameter $m$ (where $m > 0$) if its probability mass function is given by

$P(X = x) = \frac{e^{-m} m^x}{x!}, \quad x = 0, 1, 2, \dots$

where $e = 2.71828$ is the base of natural logarithms and $m$ is the mean number of occurrences.

Properties:

1. Mean and variance are equal, i.e. $E(X) = \text{Var}(X) = m$.
2. It is the limiting case of the binomial distribution when $n \to \infty$, $p \to 0$ and $np = m$ (a constant).
3. The sum of independent Poisson variates is also a Poisson variate.

Applications: It is used to model the number of telephone calls received at a switchboard per minute, the number of accidents on a road, the number of defective items in a large lot, the number of radioactive particles emitted per second, and the number of typing errors per page.

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(iii) Association of Attributes

When two or more attributes are studied together, the relationship between them is called association of attributes. Two attributes $A$ and $B$ are said to be associated if they are not independent, i.e. the presence or absence of one attribute influences the presence or absence of the other. If they are independent, they are said to have no association.

For two attributes $A$ and $B$, the data are arranged in a $2 \times 2$ contingency table:

Conditions for independence: If $A$ and $B$ are independent, the expected frequencies are obtained from the product of marginal proportions, i.e. $(AB) = \frac{(A)(B)}{N}$, etc. The observed and expected frequencies are then compared.

Measures of association:

1. Yule's coefficient of association: $Q = \frac{ad - bc}{ad + bc}$ $Q$ lies between $-1$ and $+1$. $Q = 0$ means no association, $Q = +1$ means perfect positive association and $Q = -1$ means perfect negative association.

2. Yule's coefficient of colligation: $Y = \frac{\sqrt{ad} - \sqrt{bc}}{\sqrt{ad} + \sqrt{bc}}$

3. Coefficient of contingency (for larger tables) based on the chi-square test.

Applications: Association of attributes is used in sociology, education, medicine and genetics — for example, to study whether literacy is associated with employment, or whether a particular treatment is associated with recovery.

Normal distribution

A continuous random variable $X$ is said to follow a normal distribution if its probability density function is

$$f(x)=\frac{1}{\sigma\sqrt{2\pi}}e^{-\frac{1}{2}\left(\frac{x-\mu}{\sigma}\right)^2},\qquad -\infty<x<\infty$$

where $\mu$ is the mean and $\sigma$ is the standard deviation. It is denoted by

$$X\sim N(\mu,\sigma^2)$$

The curve is bell-shaped and symmetrical about the mean. The total area under the curve is 1. For calculation, we use the standard normal variable

$$Z=\frac{X-\mu}{\sigma}\sim N(0,1)$$

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Given

$$\mu=164.2\text{ cm},\qquad \sigma=2.7\text{ cm}$$

(i) Probability that a girl’s height is more than 168 cm

$$P(X>168)=P\left(Z>\frac{168-164.2}{2.7}\right)$$ $$=P(Z>1.4074)$$

From the standard normal table,

$$P(Z<1.41)=0.9207$$

So,

$$P(Z>1.4074)=1-P(Z<1.4074)$$ $$\approx 1-0.9203=0.0797$$ $$\boxed{P(X>168)\approx 0.0797 \text{ or } 7.97\%}$$

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(ii) Lowest height of the tallest 10% girls

The tallest 10% girls have heights greater than the 90th percentile. So we need $x$ such that

$$P(X>x)=0.10$$

or,

$$P(X<x)=0.90$$

For standard normal,

$$P(Z<z)=0.90 \implies z=1.2816$$

Now,

$$x=\mu+z\sigma$$ $$x=164.2+1.2816(2.7)$$ $$x=164.2+3.4603$$ $$\boxed{x\approx 167.66\text{ cm}}$$

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Final answers

(i) $P(X>168)\approx 0.0797$ or $7.97\%$

(ii) Lowest height of the tallest 10% girls $\approx 167.66$ cm.

Given data:

$n = 8,\quad \mu_0 = 16 \text{ oz},\quad \alpha = 0.05$

Sample observations: 15.9, 16.1, 15.8, 15.7, 15.9, 16.1, 15.8, 16.0

Sample mean:

$\bar{x} = \frac{15.9+16.1+15.8+15.7+15.9+16.1+15.8+16.0}{8} = \frac{127.3}{8} = 15.9125 \text{ oz}$

Sample standard deviation:

$s = \sqrt{\frac{\sum (x_i-\bar{x})^2}{n-1}} = \sqrt{\frac{0.14875}{7}} = \sqrt{0.02125} = 0.1458 \text{ oz}$

Hypotheses:

$H_0: \mu = 16 \quad \text{(the machine fills 16 oz on average)}$ $H_1: \mu < 16 \quad \text{(the mean fill weight is less than 16 oz)}$

Test statistic (population SD unknown, so t‑test):

$t = \frac{\bar{x} - \mu_0}{s/\sqrt{n}} = \frac{15.9125 - 16}{0.1458/\sqrt{8}} = \frac{-0.0875}{0.05155} = -1.698$

Critical value: for a left‑tailed test at $\alpha = 0.05$ with $df = n-1 = 7$,

$t_{0.05,\,7} = -1.895$

Decision: Since $t_{cal} = -1.698 > -1.895$, the calculated value does not fall in the critical (rejection) region. We fail to reject $H_0$.

Conclusion: At the 5% level of significance there is not enough evidence to conclude that the mean fill weight is less than 16 ounces. The sample mean (15.9125 oz) is lower than 16 oz, but the difference is not statistically significant — it can reasonably be attributed to chance variation. The machine appears to be filling bags correctly on average.

Assumptions for Analysis of Variance (ANOVA)

1. Independence — The observations are independent of one another (random sampling / random assignment).
2. Normality — The observations in each group (treatment) follow a normal distribution.
3. Homogeneity of variance (homoscedasticity) — The population variances of all groups are equal, i.e. $\sigma_1^2 = \sigma_2^2 = \sigma_3^2$.
4. Additivity — The total variation is the sum of treatment effect and random error, i.e. the model $x_{ij} = \mu + \tau_i + e_{ij}$ holds, with $e_{ij} \sim N(0, \sigma^2)$.

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Computation

Let $k = 3$ treatments (temperatures), $n = 3$ replications, $N = 9$.

Group totals and means:

Grand total $G = 242 + 275 + 264 = 781$, so grand mean $\bar{x} = 781/9 = 86.778$.

Correction factor: $C = \frac{G^2}{N} = \frac{781^2}{9} = \frac{609961}{9} = 67773.444$

Total sum of squares: $\sum x_{ij}^2 = 80^2+81^2+81^2+93^2+90^2+92^2+87^2+89^2+88^2 = 67969$

$TSS = 67969 - 67773.444 = 195.556$

Treatment sum of squares: $SST = \frac{242^2}{3} + \frac{275^2}{3} + \frac{264^2}{3} - C = \frac{58564 + 75625 + 69696}{3} - 67773.444$ $= \frac{203885}{3} - 67773.444 = 67961.667 - 67773.444 = 188.222$

Error sum of squares: $SSE = TSS - SST = 195.556 - 188.222 = 7.333$

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ANOVA Table

Degrees of freedom: $k-1 = 2$, $N-k = 6$, $N-1 = 8$.

$F = \frac{MST}{MSE} = \frac{94.111}{1.222} = 77.0$

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Conclusion

Critical value at 5% level: $F_{0.05}(2, 6) = 5.14$.

Since calculated $F = 77.0 > 5.14$, we reject $H_0$.

Conclusion: There is sufficient evidence at the 5% level to conclude that the mean yields differ with temperature.

Let the two attributes be (A) and (B), and let (\alpha,\beta) denote their absence. The four classes are:

Two attributes are said to be positively associated when the presence of one attribute is accompanied by the presence of the other more often than expected under independence, i.e.
[ (AB) > \frac{(A)(B)}{N} ] or Yule’s coefficient (Q>0).

They are said to be negatively associated when the presence of one attribute is accompanied by the absence of the other more often than expected, i.e.
[ (AB) < \frac{(A)(B)}{N} ] or (Q<0).

If (Q=0), the attributes are independent.

Given: [ (AB)=75,\quad (A)=80,\quad (\alpha\beta)=42,\quad N=145 ]

First find the remaining frequencies: [ (A\beta)=(A)-(AB)=80-75=5 ] [ (\alpha)=N-(A)=145-80=65 ] [ (\alpha B)=(\alpha)-(\alpha\beta)=65-42=23 ]

Check: [ (AB)+(A\beta)+(\alpha B)+(\alpha\beta)=75+5+23+42=145=N ]

Yule’s coefficient of association is [ Q=\frac{(AB)(\alpha\beta)-(A\beta)(\alpha B)} {(AB)(\alpha\beta)+(A\beta)(\alpha B)} ]

Substituting: [ Q=\frac{75\times 42-5\times 23}{75\times 42+5\times 23} ] [ Q=\frac{3150-115}{3150+115} ] [ Q=\frac{3035}{3265} ] [ Q=\frac{607}{653}\approx 0.93 ]

Since (Q\approx 0.93>0), and it is close to (+1), the attributes (A) and (B) are positively associated to a high degree. That is, the presence of attribute (A) is strongly accompanied by the presence of attribute (B).

Treat the classes as continuous (0.2–0.3, 0.3–0.4, ..., 0.6–0.7), with mid-values $x = 0.25, 0.35, 0.45, 0.55, 0.65$, class width $h=0.1$ and $N=120$.

i. Histogram and frequency polygon. Draw weight (ounces) on the x-axis and frequency on the y-axis.

• Histogram: five adjacent rectangles, each 0.1 wide, with the heights shown above. The tallest bar is over 0.3–0.4.
• Frequency polygon: plot the points (mid-value, frequency) and join them by straight lines. Close the polygon at both ends by joining to the mid-values of the empty classes, $(0.15, 0)$ and $(0.75, 0)$. The polygon rises to a peak at 0.35 and has a longer tail to the right.

ii. Skewness. Take $d = (x-0.45)/0.1$:

$\bar x = 0.45 + 0.1\times\frac{-80}{120} = 0.3833$

$\sigma = 0.1\sqrt{\frac{204}{120}-\left(\frac{-80}{120}\right)^2} = 0.1\sqrt{1.7-0.4444} = 0.1121$

Mode: the modal class is 0.3–0.4, so

$\text{Mode} = 0.3 + \frac{46-30}{2(46)-30-24}\times0.1 = 0.3 + \frac{16}{38}\times0.1 = 0.3421$

$\text{Karl Pearson's coefficient: } Sk = \frac{\bar x - \text{Mode}}{\sigma} = \frac{0.3833-0.3421}{0.1121} \approx 0.37$

As a check, the moment coefficient is $\mu_3 = 0.1^3\left[-\tfrac{224}{120} + 3\left(\tfrac{80}{120}\right)\tfrac{204}{120} - 2\left(\tfrac{80}{120}\right)^3\right] \approx 9.41\times10^{-4}$, giving $\gamma_1 = \mu_3/\sigma^3 \approx 0.67$.

Interpretation: The skewness is positive. The distribution is moderately positively skewed: most pears are light (modal weight about 0.34 oz, below the mean 0.38 oz), with a tail of heavier pears to the right.

iii. Percentage of pears heavier than 0.45 oz. The value 0.45 lies in the class 0.4–0.5 (24 pears). Assuming the pears are spread evenly in the class, half of them (12) weigh more than 0.45 oz. Add the higher classes:

$12 + 14 + 6 = 32$

$\text{Percentage} = \frac{32}{120}\times100 \approx 26.7\%$

So about 26.7% of the pears weigh more than 0.45 ounces (equivalently, the cumulative frequency below 0.45 oz is $30+46+12 = 88$, i.e. 73.3%).

Positive and negative correlation

Positive correlation: Two variables are said to be positively correlated if they tend to move in the same direction, i.e. when one variable increases, the other also increases (and vice versa). The correlation coefficient $r$ is positive.

Negative correlation: Two variables are said to be negatively correlated if they tend to move in opposite directions, i.e. when one variable increases, the other decreases. The correlation coefficient $r$ is negative.

The correlation coefficient always lies between $-1$ and $+1$. Values near $+1$ indicate strong positive correlation, values near $-1$ indicate strong negative correlation, and values near $0$ indicate little or no linear correlation.

Let

• $X$ = family size
• $Y$ = number of telephone calls per week

Here,

$n=10,\quad \sum X=60,\quad \sum Y=380,\quad \sum X^2=376,\quad \sum Y^2=18646,\quad \sum XY=2488$

Also,

$\bar X=\frac{60}{10}=6,\qquad \bar Y=\frac{380}{10}=38$

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i. Correlation coefficient

Karl Pearson’s correlation coefficient is

$$r=\frac{n\sum XY-\sum X\sum Y} {\sqrt{\left[n\sum X^2-(\sum X)^2\right]\left[n\sum Y^2-(\sum Y)^2\right]}}$$

Substituting the values:

$$r=\frac{10(2488)-(60)(380)} {\sqrt{\left[10(376)-60^2\right]\left[10(18646)-380^2\right]}}$$ $$r=\frac{24880-22800} {\sqrt{(3760-3600)(186460-144400)}}$$ $$r=\frac{2080}{\sqrt{(160)(42060)}}$$ $$r=\frac{2080}{\sqrt{6729600}} =\frac{2080}{2594.147}$$ $$\boxed{r=0.8018\approx 0.802}$$

Interpretation

There is a strong positive correlation between family size and the number of telephone calls per week. As the number of persons in a household increases, the number of telephone calls per week also tends to increase.

Also,

$$r^2=(0.8018)^2\approx 0.643$$

So about 64.3% of the variation in weekly telephone calls is explained by family size.

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ii. Linear regression equation of number of calls on family size

The regression line of $Y$ on $X$ is

$$Y=a+bX$$

where

$$b=\frac{n\sum XY-\sum X\sum Y}{n\sum X^2-(\sum X)^2}$$ $$b=\frac{2080}{160}=13$$

and

$$a=\bar Y-b\bar X$$ $$a=38-13(6)=38-78=-40$$

Hence the required regression equation is

$$\boxed{Y=-40+13X}$$

or,

$$\boxed{\text{Number of calls}=-40+13(\text{family size})}$$

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iii. Estimate the number of calls for a family of size 9

Put $X=9$ in the regression equation:

$$Y=-40+13(9)$$ $$Y=-40+117$$ $$\boxed{Y=77}$$

So a family of size 9 is estimated to make about 77 telephone calls per week.

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iv. Standard error of estimate

The standard error of estimate of $Y$ on $X$ is

$$S_{YX}=\sqrt{\frac{\sum Y^2-a\sum Y-b\sum XY}{n-2}}$$

Substituting the values:

$$S_{YX}=\sqrt{\frac{18646-(-40)(380)-13(2488)}{10-2}}$$ $$S_{YX}=\sqrt{\frac{18646+15200-32344}{8}}$$ $$S_{YX}=\sqrt{\frac{1502}{8}}$$ $$S_{YX}=\sqrt{187.75}$$ $$\boxed{S_{YX}=13.70\text{ calls per week approximately}}$$

So the standard error of estimate is about 13.70 calls per week.

Let the difference be
[ d=\text{Temperature before drug}-\text{Temperature after drug} ]

Here (n=8).

[ \bar d=\frac{\sum d}{n}=\frac{4.8}{8}=0.6 ]

[ s_d=\sqrt{\frac{\sum d^2-n\bar d^2}{n-1}} =\sqrt{\frac{7.44-8(0.6)^2}{7}} =\sqrt{\frac{4.56}{7}} =0.8071 ]

Paired (t)-test:

[ H_0:\mu_d=0 \quad \text{(drug does not change body temperature)} ] [ H_1:\mu_d\ne 0 \quad \text{(drug changes body temperature)} ]

Test statistic:

[ t=\frac{\bar d}{s_d/\sqrt n} =\frac{0.6}{0.8071/\sqrt8} =\frac{0.6}{0.2854} =2.1026 ]

Degrees of freedom:

[ df=n-1=7 ]

At (5%) level of significance, two-tailed critical value:

[ t_{0.025,7}=2.365 ]

Since

[ |t|=2.1026<2.365 ]

we fail to reject (H_0).

Also, the approximate (p)-value is about (0.073), which is greater than (0.05).

Conclusion: At the (5%) level of significance, there is not enough evidence to conclude that the new drug changes body temperature.