Statistics I

(i) Constant $k$. The total probability must be 1:

$\int_0^1\!\!\int_0^1 k(2x+3y)\,dx\,dy=k\int_0^1\left[x^2+3xy\right]_0^1dy=k\int_0^1(1+3y)\,dy=k\left[y+\tfrac{3y^2}{2}\right]_0^1=\tfrac{5}{2}k=1$

so $k=\dfrac{2}{5}=0.4$ and $f(x,y)=\dfrac{2}{5}(2x+3y)$.

(ii) Conditional pdf of $X$ given $Y=y$. First find the marginal density of $Y$:

$f_Y(y)=\int_0^1\tfrac25(2x+3y)\,dx=\tfrac25\left[x^2+3xy\right]_0^1=\tfrac25(1+3y),\qquad 0\le y\le1$

Then

$f(x\mid y)=\frac{f(x,y)}{f_Y(y)}=\frac{\tfrac25(2x+3y)}{\tfrac25(1+3y)}=\frac{2x+3y}{1+3y},\qquad 0\le x\le1$

(iii) Independence. The marginal density of $X$ is

$f_X(x)=\int_0^1\tfrac25(2x+3y)\,dy=\tfrac25\left(2x+\tfrac32\right)=\tfrac{2}{5}\cdot\tfrac{4x+3}{2}$

so $f_X(x)=\dfrac{4x+3}{5}$. Now

$f_X(x)f_Y(y)=\frac{(4x+3)(2)(1+3y)}{25}\ne\tfrac25(2x+3y)=f(x,y)$

For example, at $x=y=0$ the product is $\tfrac{6}{25}=0.24$ but $f(0,0)=0$. Equivalently, $f(x\mid y)$ depends on $y$. Hence $X$ and $Y$ are not independent.

Assumptions of Pearson's correlation

• Both variables are quantitative (interval or ratio scale).
• The relationship between them is linear.
• Both variables are approximately normally distributed (bivariate normal).
• Observations are paired and independent of each other.
• There are no extreme outliers, and the spread is roughly constant (homoscedasticity).

Let $X$ = number of rooms and $Y$ = kilowatt-hours (thousands), with $n=5$.

(a)

$r=\frac{n\sum XY-\sum X\sum Y}{\sqrt{n\sum X^2-(\sum X)^2}\sqrt{n\sum Y^2-(\sum Y)^2}}=\frac{5(449)-51(42)}{\sqrt{5(547)-51^2}\sqrt{5(372)-42^2}}$

$r=\frac{2245-2142}{\sqrt{134}\sqrt{96}}=\frac{103}{11.576\times9.798}=0.908$

Interpretation: $r=0.91$ shows a strong positive linear relationship. Houses with more rooms use more electricity. Also $r^2=0.825$, so about 82.5% of the variation in electricity use is explained by the number of rooms.

(b) Regression of $Y$ on $X$:

$b_{YX}=\frac{n\sum XY-\sum X\sum Y}{n\sum X^2-(\sum X)^2}=\frac{103}{134}=0.7687$

$\bar X=10.2,\quad \bar Y=8.4,\quad a=\bar Y-b\bar X=8.4-0.7687(10.2)=0.5597$

$\hat Y=0.560+0.769X$

Interpretation: for each additional room, the electricity used rises on average by 0.769 thousand kWh (about 769 kWh).

(c) For $X=8$:

$\hat Y=0.560+0.769(8)=6.71$

The estimated consumption is about 6.71 thousand kWh (about 6710 kWh).

Conditional probability: the probability of an event $A$ given that another event $B$ has already occurred is

$P(A\mid B)=\frac{P(A\cap B)}{P(B)},\qquad P(B)>0$

Problem. The chances of solving are in the ratio $2:3:5$, so (taking the total as 10)

$P(A)=\frac{2}{10}=0.2,\quad P(B)=\frac{3}{10}=0.3,\quad P(C)=\frac{5}{10}=0.5$

Assume the students work independently.

(i) None of them solves the problem:

$P(\text{none})=(1-0.2)(1-0.3)(1-0.5)=0.8\times0.7\times0.5=0.28$

(ii) The problem is solved (at least one solves it):

$P(\text{solved})=1-P(\text{none})=1-0.28=0.72$

Correlation analysis: a statistical technique for studying the existence, direction and strength of the linear relationship between two quantitative variables. The Karl Pearson coefficient $r$ lies between $-1$ and $+1$: $r>0$ shows positive correlation, $r<0$ negative correlation and $r=0$ no linear correlation.

Let $X$ = number of rooms, $Y$ = kilowatt-hours (thousands), $n=5$.

$\sum X=50,\ \sum Y=40,\ \sum XY=426,\ \sum X^2=520,\ \sum Y^2=354$

$S_{xy}=426-\frac{50\times40}{5}=26,\quad S_{xx}=520-\frac{50^2}{5}=20,\quad S_{yy}=354-\frac{40^2}{5}=34$

(i)

$r=\frac{S_{xy}}{\sqrt{S_{xx}S_{yy}}}=\frac{26}{\sqrt{20\times34}}=\frac{26}{26.08}\approx0.997$

Interpretation: there is a very strong positive linear correlation; houses with more rooms use more electricity. Also $r^2\approx99.4\%$ of the variation in consumption is explained by the number of rooms.

(ii) Regression of $Y$ on $X$:

$b=\frac{S_{xy}}{S_{xx}}=\frac{26}{20}=1.3,\quad \bar x=10,\ \bar y=8,\quad a=8-1.3(10)=-5$

$\hat Y=-5+1.3X$

(iii) For $X=8$:

$\hat Y=-5+1.3(8)=5.4$

An 8-room house is estimated to use about 5.4 thousand kilowatt-hours.

Independent events: two events $A$ and $B$ are independent if the occurrence of one does not affect the probability of the other, i.e. $P(A\cap B)=P(A)P(B)$.

Problem: the chances of A, B, C are in the ratio $2:3:5$, so (the total of the ratio being 10)

$P(A)=\tfrac{2}{10}=0.2,\quad P(B)=\tfrac{3}{10}=0.3,\quad P(C)=\tfrac{5}{10}=0.5$

The students work independently.

(i) All of them solve the problem:

$P(A\cap B\cap C)=0.2\times0.3\times0.5=0.03$

(ii) The problem is solved if at least one of them solves it:

$P=1-P(\bar A)P(\bar B)P(\bar C)=1-(0.8)(0.7)(0.5)=1-0.28=0.72$

Absolute and relative measures of dispersion

Consistency of computers. The series with the smaller coefficient of variation, $CV = \dfrac{\sigma}{\bar{x}} \times 100$, is more consistent. Use mid-values $m = 1, 3, 5, 7, 9, 11$ and $n = 50$ for each company.

Company A ($f = 5, 16, 13, 7, 5, 4$)

$\sum fm = 5 + 48 + 65 + 49 + 45 + 44 = 256,\quad \bar{x}_A = \frac{256}{50} = 5.12$

$\sum fm^2 = 5 + 144 + 325 + 343 + 405 + 484 = 1706$

$\sigma_A^2 = \frac{1706}{50} - (5.12)^2 = 34.12 - 26.2144 = 7.9056,\quad \sigma_A = 2.812$

$CV_A = \frac{2.812}{5.12} \times 100 = 54.92\%$

Company B ($f = 2, 7, 12, 19, 9, 1$)

$\sum fm = 2 + 21 + 60 + 133 + 81 + 11 = 308,\quad \bar{x}_B = \frac{308}{50} = 6.16$

$\sum fm^2 = 2 + 63 + 300 + 931 + 729 + 121 = 2146$

$\sigma_B^2 = \frac{2146}{50} - (6.16)^2 = 42.92 - 37.9456 = 4.9744,\quad \sigma_B = 2.230$

$CV_B = \frac{2.230}{6.16} \times 100 = 36.21\%$

Since $CV_B = 36.21\% < CV_A = 54.92\%$, the computers of company B are more consistent.

(a) Shear stress depends on normal stress, so shear stress ($Y$) is the response variable and normal stress ($X$) is the explanatory variable. Fit $Y = a + bX$.

With $n = 9$: $\sum X = 230$, $\sum Y = 217$, $\sum XY = 5522$, $\sum X^2 = 5908$, $\sum Y^2 = 5273$.

$S_{xy} = 5522 - \frac{230 \times 217}{9} = 5522 - 5545.56 = -23.56$

$S_{xx} = 5908 - \frac{230^2}{9} = 5908 - 5877.78 = 30.22$

$b = \frac{S_{xy}}{S_{xx}} = \frac{-23.56}{30.22} = -0.779$

$a = \bar{Y} - b\bar{X} = 24.111 - (-0.779)(25.556) = 44.03$

Fitted line:

$\hat{Y} = 44.03 - 0.779X$

(b) The regression coefficient $b = -0.779$ means that for every 1 unit increase in normal stress, the shear resistance decreases on average by about 0.779 units. The negative sign shows an inverse relationship.

(c) $S_{yy} = 5273 - \dfrac{217^2}{9} = 5273 - 5232.11 = 40.89$

$r^2 = \frac{S_{xy}^2}{S_{xx}S_{yy}} = \frac{(-23.56)^2}{30.22 \times 40.89} = \frac{554.9}{1235.7} = 0.449$

The coefficient of determination is about $0.449$ (44.9%): about 45% of the variation in shear resistance is explained by normal stress through the linear model, and the remaining 55% is due to other factors. ($r = -0.67$, a moderate negative correlation.)

(d) For $X = 30$:

$\hat{Y} = 44.03 - 0.779 \times 30 = 44.03 - 23.38 \approx 20.65$

The predicted shear resistance is about 20.65.

(a) Binomial distribution. A random variable $X$, the number of successes in $n$ independent Bernoulli trials, each with the same probability of success $p$, follows the binomial distribution:

$P(X = x) = \binom{n}{x} p^x q^{n-x},\quad x = 0, 1, \dots, n,\quad q = 1 - p$

It is written $X \sim B(n, p)$.

Main features:

1. The experiment has a fixed number $n$ of trials.
2. Each trial has only two outcomes, success or failure.
3. The trials are independent and $p$ is constant for every trial.
4. Mean $= np$, variance $= npq$ (so variance is less than the mean), standard deviation $= \sqrt{npq}$.
5. It is symmetric when $p = 0.5$, positively skewed when $p < 0.5$ and negatively skewed when $p > 0.5$.
6. It has $n + 1$ possible values, and $\sum P(x) = (q + p)^n = 1$.
7. The mode is the integer part of $(n+1)p$.

(b) Marginal probability distribution. For a joint distribution of discrete random variables $X$ and $Y$ with joint pmf $p(x, y)$, the marginal distribution of one variable is obtained by summing the joint probability over all values of the other:

$p_X(x) = \sum_y p(x, y), \qquad p_Y(y) = \sum_x p(x, y)$

(for continuous variables $f_X(x) = \int f(x,y)\,dy$ and $f_Y(y) = \int f(x,y)\,dx$).

Properties:

1. $p_X(x) \ge 0$ and $p_Y(y) \ge 0$ for all values.
2. $\sum_x p_X(x) = 1$ and $\sum_y p_Y(y) = 1$ (total probability is 1).
3. $X$ and $Y$ are independent if and only if $p(x, y) = p_X(x)\,p_Y(y)$ for all $x, y$.
4. The marginal distributions are used to find the means, variances and probabilities of each variable separately.

The modal class is the class with the highest frequency, which is 49-54 (frequency 25).

$\text{Mode} = L + \frac{\Delta_1}{\Delta_1 + \Delta_2}\times h$

Here $L = 49$, $h = 5$, $f_1 = 25$, $f_0 = 16$ (previous), $f_2 = 5$ (next):

$\Delta_1 = f_1 - f_0 = 25 - 16 = 9,\qquad \Delta_2 = f_1 - f_2 = 25 - 5 = 20$

$\text{Mode} = 49 + \frac{9}{9 + 20}\times 5 = 49 + 1.55 = 50.55$

Interpretation: The most common thickness of the computer chips is about 50.55 nanometers.

$N = 8 + 12 + 36 + 25 + 28 + 18 + 9 + 12 + 6 = 154$. Cumulative frequencies:

$Q_3$: position $= \dfrac{3N}{4} = 115.5$. The first c.f. greater than 115.5 is 127, so $Q_3 = 35$.

$D_6$: position $= \dfrac{6N}{10} = 92.4$. The first c.f. greater than 92.4 is 109, so $D_6 = 30$.

$P_{80}$: position $= \dfrac{80N}{100} = 123.2$. The first c.f. greater than 123.2 is 127, so $P_{80} = 35$.

Interpretation

• $Q_3 = 35$: 75% of the persons have a respiratory rate of 35 or less, and 25% have a rate of 35 or more.
• $D_6 = 30$: 60% of the persons have a respiratory rate of 30 or less.
• $P_{80} = 35$: 80% of the persons have a respiratory rate of 35 or less, and the remaining 20% have a higher rate.

Random variable. A random variable is a function that assigns a real number to each outcome of the sample space of a random experiment. It is discrete if it takes countable values and continuous if it takes any value in an interval.

(a) Marginal pmf. Add the joint probabilities along rows and columns.

For $X$ (row sums):

$P(X=0) = \frac{1+2+2+3}{32} = \frac{8}{32} = \frac14$

$P(X=1) = \frac{1}{16}+\frac{1}{16}+4\cdot\frac18 = \frac18 + \frac12 = \frac58$

$P(X=2) = \frac{2}{32} + 4\cdot\frac{1}{64} = \frac{1}{16}+\frac{1}{16} = \frac18$

For $Y$ (column sums):

Check: $\tfrac{6+6+11+13+13+15}{64} = 1$ and $\tfrac14+\tfrac58+\tfrac18 = 1$.

(b)

$P(X \le 1, Y = 2) = P(0,2) + P(1,2) = 0 + \frac{1}{16} = \frac{1}{16}$

(c)

$P(X \le 1) = P(X=0) + P(X=1) = \frac14 + \frac58 = \frac78$

(i) Constant $k$. The total probability must be 1:

$\int_0^\infty\!\!\int_0^\infty k e^{-(x+y)}\,dx\,dy = k\left(\int_0^\infty e^{-x}dx\right)\left(\int_0^\infty e^{-y}dy\right) = k(1)(1) = k$

So $k = 1$ and $f(x,y) = e^{-x}e^{-y}$.

(ii) Conditional pdf of $X$ given $Y$. The marginal of $Y$ is

$f_Y(y) = \int_0^\infty e^{-(x+y)}dx = e^{-y},\quad y > 0$

$f(x \mid y) = \frac{f(x,y)}{f_Y(y)} = \frac{e^{-(x+y)}}{e^{-y}} = e^{-x},\quad x > 0$

This equals the marginal of $X$, so $X$ and $Y$ are independent.

(iii) $\operatorname{Var}(3X + 2Y)$. $X$ and $Y$ are independent exponential variables with parameter 1, so $\operatorname{Var}(X) = \operatorname{Var}(Y) = 1$ and covariance is 0:

$\operatorname{Var}(3X + 2Y) = 9\operatorname{Var}(X) + 4\operatorname{Var}(Y) = 9(1) + 4(1) = 13$

Let $X$ be the resistance, $X \sim N(\mu = 40, \sigma = 2)$ and $Z = \dfrac{X - 40}{2}$.

(a) Resistance exceeding 43 ohms

$z = \frac{43 - 40}{2} = 1.5$

$P(X > 43) = P(Z > 1.5) = 1 - 0.9332 = 0.0668$

About 6.68% of the resistors have a resistance exceeding 43 ohms.

(b) Resistance between 30 and 45 ohms

$z_1 = \frac{30 - 40}{2} = -5,\qquad z_2 = \frac{45 - 40}{2} = 2.5$

$P(30 < X < 45) = P(-5 < Z < 2.5) = \Phi(2.5) - \Phi(-5) = 0.9938 - 0.0000 = 0.9938$

About 99.38% of the resistors have a resistance between 30 and 45 ohms.

(a) Ranking the data. Rank each variable from the smallest value, giving tied values the average of their ranks.

(b) Tie correction. Anger has two ties of size 2 (values 5 and 13) and Vigor has one tie of size 3 (value 19). Each tie group adds $\dfrac{m^3-m}{12}$ to $\sum d^2$:

$\frac{2^3-2}{12}+\frac{2^3-2}{12}+\frac{3^3-3}{12}=0.5+0.5+2=3$

(c) Rank correlation. With $n=8$:

$r_s=1-\frac{6\left(\sum d^2+3\right)}{n(n^2-1)}=1-\frac{6(113)}{8(63)}=1-\frac{678}{504}=-0.345$

Interpretation. $r_s\approx-0.35$ shows a weak negative rank correlation. Players with higher anger scores tend to have somewhat lower vigor scores, but the relationship is weak (for $n=8$ it is not significant at 5%).

Step 1: class boundaries. The classes are 23-27, 28-32, ... so the true boundaries are 22.5-27.5, 27.5-32.5, and so on. The class width is $h=5$.

$N=55$. The percentile coefficient of kurtosis is

$K=\frac{Q_3-Q_1}{2\,(P_{90}-P_{10})}=\frac{Q_D}{P_{90}-P_{10}}$

Step 2: required partition values, using $X_p=L+\dfrac{pN-cf}{f}\,h$.

• $Q_1$: $N/4=13.75$, in class 22.5-27.5: $Q_1=22.5+\dfrac{13.75-0}{22}(5)=25.625$
• $Q_3$: $3N/4=41.25$, in class 32.5-37.5: $Q_3=32.5+\dfrac{41.25-38}{9}(5)=34.306$
• $P_{10}$: $5.5$, in class 22.5-27.5: $P_{10}=22.5+\dfrac{5.5}{22}(5)=23.75$
• $P_{90}$: $49.5$, in class 37.5-42.5: $P_{90}=37.5+\dfrac{49.5-47}{4}(5)=40.625$

Step 3: coefficient.

$K=\frac{34.306-25.625}{2\,(40.625-23.75)}=\frac{8.681}{33.75}=0.257$

Interpretation. The value for a normal distribution is $K=0.263$. Since $0.257<0.263$, the distribution is slightly leptokurtic (a little more peaked than normal), and it is very close to mesokurtic. The wage distribution is also positively skewed, with most workers in the lowest wage classes.

Properties of the Poisson distribution

1. It is a discrete distribution for the number of rare events in a fixed interval of time, area or volume: $P(X=x)=\dfrac{e^{-\lambda}\lambda^x}{x!}$, $x=0,1,2,\dots$
2. It has a single parameter $\lambda>0$, the average number of occurrences.
3. Mean $=$ variance $=\lambda$. The standard deviation is $\sqrt{\lambda}$.
4. It is the limit of the binomial distribution as $n\to\infty$, $p\to0$ with $np=\lambda$ fixed.
5. It is positively skewed; the skewness is $1/\sqrt\lambda$ and falls as $\lambda$ increases. It is approximately normal for large $\lambda$.
6. Events occur independently, and the probability of two or more events in a very small interval is negligible.
7. The recurrence $P(x+1)=\dfrac{\lambda}{x+1}P(x)$ holds. The sum of independent Poisson variables is Poisson with parameter $\lambda_1+\lambda_2$.

Fitting the distribution

$N=\sum f=71+112+117+57+27+11+3+1=399$ and $\sum fx=112+234+171+108+55+18+7=705$.

$\lambda=\bar x=\frac{705}{399}=1.767$

$e^{-1.767}=0.17086$. Expected frequency $=N\cdot P(x)$ with $P(x+1)=\dfrac{\lambda}{x+1}P(x)$:

The fitted Poisson frequencies agree reasonably with the observed ones.

Primary data are data collected afresh by the investigator, directly from the original source, for the specific purpose of the study at hand. They are collected by methods such as surveys with questionnaires, interviews, observation, experiments and focus groups. Example: a researcher interviews 200 students of a college to find their daily study hours.

Secondary data are data that were already collected by someone else for some other purpose and are now used by the investigator. They come from published sources (census reports, Nepal Rastra Bank and Ministry reports, journals, books, newspapers) and unpublished sources (office records, theses). Example: using the National Population Census report to study literacy rates.

Differences

Primary data are preferred where precision and relevance matter and resources permit; secondary data are used when time or money is limited, but their reliability, suitability and adequacy should be checked first.

Sampling. Sampling is the process of selecting a part (a sample) of the units of a population so that conclusions about the whole population can be drawn from it. It saves time and cost, and it is the only possible method when testing destroys the unit (for example testing the life of bulbs) or the population is infinite or very large.

Non-probability sampling. In non-probability sampling, units are chosen by the judgement or convenience of the investigator and not by random chance. The probability of any unit being selected is unknown, so sampling error cannot be measured.

Main types

1. Convenience sampling. Units that are easiest to reach are chosen, for example interviewing the first 50 students met in the college canteen.
2. Judgement (purposive) sampling. The investigator uses personal judgement to select units that seem most representative or useful, for example choosing typical villages for a study.
3. Quota sampling. The population is divided into groups (such as sex or age) and a fixed quota is filled from each group by non-random choice.
4. Snowball sampling. Existing respondents recommend further respondents; it is used for hidden populations, such as drug users.

Merits

• It is quick, cheap and simple to carry out.
• It needs no sampling frame.
• It is useful for exploratory or pilot studies and for rare or hard-to-reach groups.
• Expert judgement can include the most informative units.

Demerits

• It is subject to personal bias, so the sample may not be representative.
• Sampling error cannot be calculated, so results cannot be generalised with a known confidence level.
• Statistical inference (confidence intervals, tests) is not strictly valid.
• Results depend heavily on the skill of the investigator.

Role of measures of dispersion. An average gives only the central value of a distribution; it does not show how scattered the observations are. Measures of dispersion (range, quartile deviation, mean deviation, standard deviation, coefficient of variation):

• show how well the average represents the data (a small dispersion means the average is reliable);
• allow comparison of the variability or consistency of two or more series;
• help in quality control by showing the uniformity of products;
• are the basis of further techniques such as correlation, regression, sampling and testing of hypotheses.

Comparing the two companies. Both have the same unit (nanometer), but their means differ, so the coefficient of variation $CV=\dfrac{\sigma}{\bar x}\times100$ is the suitable tool.

Company A ($N=100$)

$\bar x_A=\frac{1800}{100}=18,\qquad \sigma_A=\sqrt{\frac{38100}{100}-18^2}=\sqrt{57}=7.55$

$CV_A=\frac{7.55}{18}\times100=41.94\%$

Company B ($N=100$)

$\bar x_B=\frac{1700}{100}=17,\qquad \sigma_B=\sqrt{340-289}=\sqrt{51}=7.14$

$CV_B=\frac{7.14}{17}\times100=42.01\%$

Conclusion. $CV_A=41.94\%$ is slightly smaller than $CV_B=42.01\%$, so Company A's chips are marginally more uniform (consistent) in thickness. The difference is very small, so the two companies are almost equally consistent. Company A also has the larger mean thickness (18 against 17 nm).

(a) Variables and fitted line. Electric power consumed (BTU) depends on the ambient temperature, so power ($Y$) is the response variable and temperature ($X$) is the explanatory variable.

With $n=8$: $\sum X=401$, $\sum Y=2301$, $\sum X^2=22495$, $\sum XY=118652$, $\sum Y^2=666509$.

$S_{xx}=22495-\frac{401^2}{8}=2394.875,\qquad S_{xy}=118652-\frac{(401)(2301)}{8}=3314.375$

$b=\frac{S_{xy}}{S_{xx}}=\frac{3314.375}{2394.875}=1.384$

$\bar x=50.125,\quad \bar y=287.625,\qquad a=\bar y-b\bar x=287.625-1.384(50.125)=218.25$

The fitted regression line is

$\hat Y=218.25+1.384\,X$

(b) Interpretation of $b$. The slope $b=1.384$ means that for each 1°F rise in ambient temperature, the electric power consumed by the plant increases by about 1.38 BTU on average. The intercept 218.25 is the predicted consumption at 0°F.

(c) Coefficient of determination. $S_{yy}=666509-\dfrac{2301^2}{8}=4683.875$.

$r^2=\frac{S_{xy}^2}{S_{xx}S_{yy}}=\frac{(3314.375)^2}{(2394.875)(4683.875)}=0.979$

About 97.9% of the variation in power consumption is explained by the linear relationship with ambient temperature; only about 2.1% is due to other factors. The fit is very good ($r=+0.99$).

(d) Prediction at $X=65$°F.

$\hat Y=218.25+1.384(65)=308.2\ \text{BTU}$

The power consumption is predicted to be about 308 BTU. Since 65°F lies within the observed range (27 to 74°F), this is a safe interpolation.

(a) Normal distribution. A continuous random variable $X$ follows a normal distribution with mean $\mu$ and standard deviation $\sigma$ 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 $\pi=3.1416$ and $e=2.7183$. We write $X\sim N(\mu,\sigma^2)$. If $Z=\dfrac{X-\mu}{\sigma}$ then $Z\sim N(0,1)$, the standard normal variable.

Main characteristics

1. The curve is bell-shaped and symmetrical about $x=\mu$.
2. Mean, median and mode are all equal to $\mu$.
3. It is unimodal with maximum ordinate $\dfrac{1}{\sigma\sqrt{2\pi}}$ at $x=\mu$.
4. The curve is asymptotic to the $x$-axis; the tails never touch it.
5. Skewness is $0$ and kurtosis is $3$ (mesokurtic).
6. The points of inflection are at $x=\mu\pm\sigma$.
7. The total area under the curve is 1. About 68.27% of the area lies within $\mu\pm\sigma$, 95.45% within $\mu\pm2\sigma$ and 99.73% within $\mu\pm3\sigma$.
8. Quartiles are $Q_1,Q_3=\mu\mp0.6745\sigma$; the quartile deviation is $0.6745\sigma$ and the mean deviation is $0.7979\sigma$.
9. It is completely determined by the two parameters $\mu$ and $\sigma$. A linear combination of independent normal variables is also normal.

(b) Probability density function. For a continuous random variable $X$, the probability density function (pdf) $f(x)$ is a function such that the probability that $X$ lies in an interval is the area under $f$ over that interval:

$P(a\le X\le b)=\int_a^b f(x)\,dx$

Properties

1. $f(x)\ge0$ for all $x$.
2. $\displaystyle\int_{-\infty}^{\infty}f(x)\,dx=1$ (the total area is 1).
3. $P(X=c)=0$ for any single point, so $P(a\le X\le b)=P(a<X<b)$.
4. The cumulative distribution function is $F(x)=\int_{-\infty}^{x}f(t)\,dt$, and $f(x)=\dfrac{dF(x)}{dx}$ wherever the derivative exists.
5. The mean is $E(X)=\int xf(x)\,dx$ and the variance is $E(X^2)-[E(X)]^2$.

Let the missing frequencies be $f_2$ (class 10-20) and $f_4$ (class 30-40).

Total frequency: $4+f_2+10+f_4+10=50\Rightarrow f_2+f_4=26$ ... (i)

Mean $=30.2$, so $\sum fm=30.2\times50=1510$:

$$20+15f_2+250+35f_4+450=1510\ \Rightarrow\ 15f_2+35f_4=790\ \Rightarrow\ 3f_2+7f_4=158$$

... (ii)

From (i), $f_2=26-f_4$. Substituting in (ii):

$3(26-f_4)+7f_4=158\Rightarrow 78+4f_4=158\Rightarrow f_4=20,\qquad f_2=6$

Check: $4+6+10+20+10=50$ and $\sum fm=20+90+250+700+450=1510$, giving mean $30.2$.

The missing frequencies are 6 (class 10-20) and 20 (class 30-40).

Power failure time is a discrete variable, so the cumulative frequency is used. Here $N=33$.

For a discrete series the required partition value is the value of $x$ whose cumulative frequency first reaches or exceeds the position given by $\dfrac{(N+1)k}{m}$.

$Q_3$: position $=\dfrac{3(N+1)}{4}=\dfrac{3(34)}{4}=25.5$. The first cf above 25.5 is 28, so $Q_3=26$.

$D_2$: position $=\dfrac{2(N+1)}{10}=\dfrac{2(34)}{10}=6.8$. The first cf above 6.8 is 7, so $D_2=23$.

$P_{40}$: position $=\dfrac{40(N+1)}{100}=\dfrac{40(34)}{100}=13.6$. The first cf above 13.6 is 14, so $P_{40}=24$.

Interpretation

• $Q_3=26$: 75% of the power failures last 26 minutes or less, and 25% last longer.
• $D_2=23$: 20% of the power failures last 23 minutes or less.
• $P_{40}=24$: 40% of the power failures last 24 minutes or less, and 60% last longer.

Let $D$ be the event that the product is defective. The priors are $P(P_1)=0.30$, $P(P_2)=0.20$, $P(P_3)=0.50$.

Total probability of a defective product

$P(D)=\sum P(P_j)P(D\mid P_j)=0.30(0.01)+0.20(0.03)+0.50(0.02)=0.003+0.006+0.010=0.019$

Bayes' theorem gives $P(P_j\mid D)=\dfrac{P(P_j)P(D\mid P_j)}{P(D)}$:

$P(P_1\mid D)=\frac{0.003}{0.019}=0.158$

$P(P_2\mid D)=\frac{0.006}{0.019}=0.316$

$P(P_3\mid D)=\frac{0.010}{0.019}=0.526$

The three posterior probabilities sum to 1 ($0.158+0.316+0.526=1$).

Conclusion. The highest posterior probability is for plan 3 (0.526), so plan 3 was most likely used and is most likely responsible for the defective product. This is because plan 3 is used for half of all products, even though its defect rate (0.02) is lower than that of plan 2.