Statistics and Probability

The lower quartile ($Q_1$) divides the data so that $25\%$ of the values lie below it.

Total outcomes $= 6 \times 2 = 12$.

Prime numbers on a dice are $2, 3, 5$, so $P(\text{prime}) = \dfrac{3}{6} = \dfrac{1}{2}$, and $P(\text{head}) = \dfrac{1}{2}$.

The two events are independent, so

$P(\text{prime and head}) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}.$

Odd numbers on a dice: $1,3,5$, so $P(\text{odd})=\dfrac36=\dfrac12$.

$P(\text{head})=\dfrac12$.

The two events are independent, so

$P(\text{odd and head})=\frac12\times\frac12=\frac14.$

Number of students $= \dfrac{\text{sum of weights}}{\text{average weight}} = \dfrac{540}{45} = 12$.

There are $12$ students.

Number of students $= \dfrac{\text{sum of weights}}{\text{average weight}} = \dfrac{480}{40} = 12$.

There are $12$ students.

Total outcomes $= 6 \times 2 = 12$.

Even numbers on a dice are $2, 4, 6$, so $P(\text{even}) = \dfrac{3}{6} = \dfrac{1}{2}$, and $P(\text{head}) = \dfrac{1}{2}$.

The two events are independent, so

$P(\text{even and head}) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}.$

$\text{Mean} = \frac{\sum fm}{N} \Rightarrow 50 = \frac{1850}{N} \Rightarrow N = \frac{1850}{50} = 37$

There are 37 students.

Prime numbers on a dice: $2, 3, 5$, so $P(\text{prime}) = \dfrac{3}{6} = \dfrac{1}{2}$. For the coin, $P(\text{tail}) = \dfrac{1}{2}$.

The events are independent, so

$P(\text{prime and tail}) = \frac{1}{2}\times\frac{1}{2} = \frac{1}{4}$

$\text{Mean}=\frac{\sum fm}{N} \Rightarrow 40=\frac{1200}{N} \Rightarrow N=\frac{1200}{40}=30.$

The number of students is 30.

Mean $= \dfrac{\sum fm}{N}$, so $30 = \dfrac{1800}{N}$.

$N = \frac{1800}{30} = \textbf{60 students}$

Mean $= \dfrac{\sum fm}{N}$, so

Answer: 15 students

Dice: even numbers are 2, 4, 6, so $P(\text{even}) = \dfrac36 = \dfrac12$.

Coin: $P(\text{tail}) = \dfrac12$.

The two events are independent, so

Answer: $\dfrac14$

Mean $= \dfrac{\sum fm}{N}$, so

Answer: 80 students

Mean $= \dfrac{\sum fm}{N}$

$42 = \frac{504}{N} \Rightarrow N = \frac{504}{42} = 12$

There are $12$ students.

A deck has $52$ cards with $4$ queens and $48$ non-queens. Let $Q$ = queen and $Q'$ = not a queen.

First draw: $P(Q) = \dfrac{4}{52} = \dfrac{1}{13}$, $P(Q') = \dfrac{48}{52} = \dfrac{12}{13}$.

Second draw (one card is already removed, so $51$ remain):

• After $Q$: $P(Q) = \dfrac{3}{51} = \dfrac{1}{17}$, $P(Q') = \dfrac{48}{51} = \dfrac{16}{17}$.
• After $Q'$: $P(Q) = \dfrac{4}{51}$, $P(Q') = \dfrac{47}{51}$.

                 1st draw            2nd draw           Outcome  Probability
                                  1/17 / Q  ----------- QQ     1/13 x 1/17 = 1/221
              1/13 / Q  ---------<
        Start <                   16/17 \ Q' ---------- QQ'    1/13 x 16/17 = 16/221
              12/13 \ Q'  --------<
                                  4/51 / Q  ----------- Q'Q    12/13 x 4/51 = 16/221
                                  47/51 \ Q' ---------- Q'Q'   12/13 x 47/51 = 188/221

Check: $\dfrac{1 + 16 + 16 + 188}{221} = 1$.

So $P(QQ) = \dfrac{1}{221}$, $P(QQ') = \dfrac{16}{221}$, $P(Q'Q) = \dfrac{16}{221}$, $P(Q'Q') = \dfrac{188}{221}$.

(a) Two events are independent if the occurrence of one does not affect the probability of the other.

(b) R = red, W = white. Total = 6 balls; without replacement.

Start
+-- R (2/6)
|       +-- R (1/5)  ->  RR = 1/15
|       +-- W (4/5)  ->  RW = 4/15
+-- W (4/6)
        +-- R (2/5)  ->  WR = 4/15
        +-- W (3/5)  ->  WW = 2/5

(c) $P(RR)=\frac26\times\frac15=\frac{2}{30}=\frac1{15}$

(d) $P(\text{different})=P(RW)+P(WR)=\frac26\times\frac45+\frac46\times\frac25=\frac{8}{30}+\frac{8}{30}=\frac{16}{30}=\frac{8}{15}$

$\frac{8}{15}-\frac{1}{15}=\frac{7}{15}$

The probability of different colours is more than that of both red by $\frac{7}{15}$.

(a) For independent events, $P(A\cap B)=P(A)\times P(B)$.

(b) R = red, W = white. Total = 5 balls.

Start
+-- R (2/5)
|       +-- R (1/4)  ->  RR = 1/10
|       +-- W (3/4)  ->  RW = 3/10
+-- W (3/5)
        +-- R (2/4)  ->  WR = 3/10
        +-- W (2/4)  ->  WW = 3/10

(c) $P(RR)=\frac{2}{5}\times\frac{1}{4}=\frac{2}{20}=\frac{1}{10}$

(d) $P(\text{same colour})=P(RR)+P(WW)=\frac{2}{20}+\frac{3}{5}\times\frac{2}{4}=\frac{2}{20}+\frac{6}{20}=\frac{8}{20}=\frac{2}{5}$

$P(\text{different colour})=1-\frac{2}{5}=\frac{3}{5}$

The probability of getting different colours ($\frac{3}{5}$) is greater than that of the same colour ($\frac{2}{5}$), by $\frac{1}{5}$.

(a) For independent events, $P(A\cap B)=P(A)\times P(B)$.

(b) R = red, B = black, W = white. Without replacement, so the colour drawn first cannot be drawn again.

Start
+-- R (1/3)
|       +-- B (1/2)  ->  RB = 1/6
|       +-- W (1/2)  ->  RW = 1/6
+-- B (1/3)
|       +-- R (1/2)  ->  BR = 1/6
|       +-- W (1/2)  ->  BW = 1/6
+-- W (1/3)
        +-- R (1/2)  ->  WR = 1/6
        +-- B (1/2)  ->  WB = 1/6

(c) A red and a black ball can come as RB or BR.

$P(RB)=\frac13\times\frac12=\frac16$, $P(BR)=\frac13\times\frac12=\frac16$

$P=\frac16+\frac16=\frac13$

(d) No. There is only one ball of each colour and the balls are drawn without replacement, so two balls of the same colour cannot be drawn. The probability is $0$ (impossible event).

$N = 20$, so $\dfrac{N}{2} = 10$. The first class whose c.f. is at least $10$ is $30 - 40$, so this is the median class.

$L = 30$, $c.f. = 9$ (before the median class), $f = 5$, $h = 10$.

$\text{Median} = L + \frac{\frac{N}{2} - c.f.}{f} \times h = 30 + \frac{10 - 9}{5} \times 10 = 30 + 2 = 32$

The median is $32$.

$N=30$, so $\dfrac N2=15$.

The first cumulative frequency that is not less than $15$ is $15$, so the median class is $50$ - $60$.

Here $L=50$, $cf=7$ (before the median class), $f=8$, $i=10$.

$\text{Median}=L+\frac{\frac N2-cf}{f}\times i=50+\frac{15-7}{8}\times10=50+10=\mathbf{60}.$

$N = 30$, so $\dfrac{N}{2} = 15$. The first c.f. greater than $15$ is $18$, so the median class is $30 - 40$.

$L = 30$, $c.f. = 12$, $f = 6$, $h = 10$.

$\text{Median} = L + \frac{\frac{N}{2} - c.f.}{f}\times h = 30 + \frac{15-12}{6}\times 10 = 30 + 5 = \textbf{35}$

$N = 20$, so $\dfrac N2 = 10$. The first $cf$ greater than 10 is 14, so the median class is $40 - 50$.

Here $L = 40$, $cf = 7$ (before the median class), $f = 7$, $i = 10$.

Answer: Median $\approx 44.29$

(a) For independent events, $P(B\cap R)=P(B)\times P(R)$. (Here the draws are without replacement, so for the actual experiment the second probability is conditional: $P(B\cap R)=P(B)\times P(R\mid B)$.)

(b) B = black, R = red. Total balls = 11. Without replacement, the second draw is from 10 balls.

Start
+-- B (7/11)
|       +-- B (6/10)  ->  BB = 21/55
|       +-- R (4/10)  ->  BR = 14/55
+-- R (4/11)
        +-- B (7/10)  ->  RB = 14/55
        +-- R (3/10)  ->  RR = 6/55

(c) $P(BB)=\frac{7}{11}\times\frac{6}{10}=\frac{21}{55}$ .

(d) $P(RR)=\frac{4}{11}\times\frac{3}{10}=\frac{6}{55}$ .

$P(BB)-P(RR)=\frac{21}{55}-\frac{6}{55}=\frac{3}{11}$

So the probability of both red balls is less than that of both black balls by $\frac{3}{11}$.

(a) For independent events A and B, the multiplication law is $P(A\cap B)=P(A)\times P(B)$ (in general, $P(A\cap B)=P(A)\times P(B\mid A)$).

(b) R = red, W = white. Total = 11 marbles.

Start
+-- R (6/11)
|       +-- R (5/10)  ->  RR = 3/11
|       +-- W (5/10)  ->  RW = 3/11
+-- W (5/11)
        +-- R (6/10)  ->  WR = 3/11
        +-- W (4/10)  ->  WW = 2/11

(c) $P(\text{same colour})=P(RR)+P(WW)=\frac{6}{11}\times\frac{5}{10}+\frac{5}{11}\times\frac{4}{10}=\frac{30}{110}+\frac{20}{110}=\frac{50}{110}=\frac{5}{11}$

(d) $P(\text{different colour})=1-\frac{5}{11}=\frac{6}{11}$

Since $\frac{6}{11}>\frac{5}{11}$, the probability of getting marbles of different colours is greater than that of the same colour, by $\frac{1}{11}$.

(a) For independent events, $P(A\cap B)=P(A)\times P(B)$.

(b) G = green, B = blue. Total = 12 balls; with replacement.

Start
+-- G (5/12)
|       +-- G (5/12)  ->  GG = 25/144
|       +-- B (7/12)  ->  GB = 35/144
+-- B (7/12)
        +-- G (5/12)  ->  BG = 35/144
        +-- B (7/12)  ->  BB = 49/144

(c) $P(BB)=\frac{7}{12}\times\frac{7}{12}=\frac{49}{144}$

(d) $P(\text{same})=\frac{25}{144}+\frac{49}{144}=\frac{74}{144}$

$P(\text{different})=\frac{35}{144}+\frac{35}{144}=\frac{70}{144}$

Ratio $=74:70=37:35$.

(a) For independent events, $P(A\cap B)=P(A)\times P(B)$.

(b) W = white, B = black. Total = 16 balls; with replacement.

Start
+-- W (6/16)
|       +-- W (6/16)  ->  WW = 9/64
|       +-- B (10/16)  ->  WB = 15/64
+-- B (10/16)
        +-- W (6/16)  ->  BW = 15/64
        +-- B (10/16)  ->  BB = 25/64

(c) $P(\text{same})=P(WW)+P(BB)=\frac{6}{16}\times\frac{6}{16}+\frac{10}{16}\times\frac{10}{16}=\frac{36}{256}+\frac{100}{256}=\frac{136}{256}=\frac{17}{32}$

(d) $P(\text{different})=\frac{6}{16}\times\frac{10}{16}\times2=\frac{120}{256}=\frac{15}{32}$, $P(WW)=\frac{36}{256}=\frac{9}{64}$

Difference $=\frac{120-36}{256}=\frac{84}{256}=\frac{21}{64}$

The probability of different colours is more than that of both white by $\frac{21}{64}$.

(a) For independent events, $P(A\cap B)=P(A)\times P(B)$.

(b) W = white, B = black. Total = 12 balls; with replacement.

Start
+-- W (4/12)
|       +-- W (4/12)  ->  WW = 1/9
|       +-- B (8/12)  ->  WB = 2/9
+-- B (8/12)
        +-- W (4/12)  ->  BW = 2/9
        +-- B (8/12)  ->  BB = 4/9

(c) $P(\text{same})=P(WW)+P(BB)=\frac{4}{12}\times\frac{4}{12}+\frac{8}{12}\times\frac{8}{12}=\frac{16}{144}+\frac{64}{144}=\frac{80}{144}=\frac59$

(d) $P(\text{different})=\frac{4}{12}\times\frac{8}{12}\times2=\frac{64}{144}=\frac49$

Yes, they are comparable: both are probabilities of events of the same experiment, with $\frac59+\frac49=1$ (complementary events). The probability of the same colour $\left(\frac59\right)$ is greater than that of different colours $\left(\frac49\right)$ by $\frac19$.

(a) For independent events A and B: $P(A\cap B)=P(A)\times P(B)$.

(b) F = face card (12), N = not a face card (40); no replacement.

Start
+-- F (12/52)
|       +-- F (11/51)  ->  FF = 11/221
|       +-- N (40/51)  ->  FN = 40/221
+-- N (40/52)
        +-- F (12/51)  ->  NF = 40/221
        +-- N (39/51)  ->  NN = 10/17

(c) $P(FF)=\frac{12}{52}\times\frac{11}{51}=\frac{11}{221}$

(d) Total probability $=1$.

$1-\frac{11}{221}=\frac{210}{221}$

The probability of both face cards is less than the total probability by $\frac{210}{221}$.

Taking $L$, $f$ and $c.f.$ for the class that contains the first quartile (the class in which the $\frac{N}{4}$th item lies):

$Q_1 = L + \frac{\frac{N}{4} - c.f.}{f}\times i$

$N = 20$, so $\dfrac{N}{2} = 10$. The first class whose c.f. is at least $10$ is $40 - 60$, so this is the median class.

$L = 40$, $c.f. = 5$ (before the median class), $f = 5$, $h = 20$.

$\text{Median} = L + \frac{\frac{N}{2} - c.f.}{f} \times h = 40 + \frac{10 - 5}{5} \times 20 = 40 + 20 = 60$

The median mark is $60$.

$N = 24$, so $\dfrac{N}{2} = 12$. The first c.f. exceeding 12 is 16, so the median class is $25 - 35$.

Here $L = 25$, $c.f. = 8$, $f = 8$, $h = 10$.

$\text{Median} = L + \frac{\frac{N}{2} - c.f.}{f} \times h = 25 + \frac{12 - 8}{8} \times 10 = 25 + 5 = 30$

The median is $30$.

Sample space: $6,7,\dots,30$, so $n(S)=30-6+1=25$.

Divisible by 7: $7,14,21,28$, so 4 numbers.

Divisible by 8: $8,16,24$, so 3 numbers.

No number is divisible by both (the first common multiple is 56), so the events are mutually exclusive.

$P(7\text{ or }8)=\frac{4}{25}+\frac{3}{25}=\mathbf{\frac{7}{25}}$

Total outcomes $= 30$.

Divisible by 5: $\{5,10,15,20,25,30\}$, so $n(A)=6$. Divisible by 7: $\{7,14,21,28\}$, so $n(B)=4$. Divisible by both 5 and 7 (that is, 35): none, so $n(A\cap B)=0$.

$n(A\cup B) = 6+4-0 = 10$

$P = \frac{10}{30} = \frac13$

(a) For independent events, $P(A\text{ and }B)=P(A\cap B)=P(A)\times P(B)$.

(b) Total balls $=12$. Without replacement:

First draw        Second draw        Outcome   Probability
                 /  W  4/11   ------   WW      5/12 x 4/11 = 20/132
        W 5/12 --
       /         \  B  7/11   ------   WB      5/12 x 7/11 = 35/132
 Start
       \         /  W  5/11   ------   BW      7/12 x 5/11 = 35/132
        B 7/12 --
                 \  B  6/11   ------   BB      7/12 x 6/11 = 42/132

(c) $P(BB)=\frac7{12}\times\frac6{11}=\frac{42}{132}=\frac7{22}$

(d) $P(WB)=\frac5{12}\times\frac7{11}=\frac{35}{132},\qquad P(BW)=\frac7{12}\times\frac5{11}=\frac{35}{132}$ Yes, they are equal. The two probabilities are products of the same numbers ($5, 7$ in the numerators and $12, 11$ in the denominators) taken in a different order, so they are equal even though the balls are not replaced.

(a) Mid-values are 10, 30, 50, 70, 90.

Mean $=\dfrac{\Sigma fm}{N}$, so

$56=\frac{820+50x}{14+x}$

$784+56x=820+50x$

$6x=36$

$x=6$

(b) With $x=6$ the frequencies are 2, 3, 6, 5, 4. The highest frequency is 6, so the modal class is 40-60.

(c) Frequencies 2, 3, 6, 5, 4; $N=20$.

Here $N=\Sigma f=20$.

$\frac{N}{4}=5$. The first c.f. not less than 5 is 5, so the first quartile class is $20-40$.

$L=20,\ c.f.=2,\ f=3,\ i=20$

$Q_1=L+\frac{\frac{N}{4}-c.f.}{f}\times i=20+\frac{5-2}{3}\times 20 = 40$

The first quartile is 40.

(d) Students securing 40 or above: $6+5+4=15$. Students securing less than 40: $2+3=5$.

So the number of students securing 40 or above is three times (ratio $15:5=3:1$) the number securing less than 40; it is 10 more.

(a) For mutually exclusive events A and B, $P(A\cup B)=P(A)+P(B)$.

(b) W = white, B = black. Total = 10. With replacement, both draws are from 10 balls.

Start
+-- W (6/10)
|       +-- W (6/10)  ->  WW = 9/25
|       +-- B (4/10)  ->  WB = 6/25
+-- B (4/10)
        +-- W (6/10)  ->  BW = 6/25
        +-- B (4/10)  ->  BB = 4/25

(c) $P(\text{same colour})=P(WW)+P(BB)=\frac{6}{10}\times\frac{6}{10}+\frac{4}{10}\times\frac{4}{10}=\frac{36}{100}+\frac{16}{100}=\frac{13}{25}=0.52$

(d) $P(\text{different colours})=P(WB)+P(BW)=\frac{24}{100}+\frac{24}{100}=\frac{48}{100}=\frac{12}{25}$

Ratio $=\frac{52}{100}:\frac{48}{100}=52:48=13:12$.

(a) $L$ represents the lower limit of the median class.

(b) Here $N=\Sigma f=30$.

$\frac{N}{2}=15$. The first c.f. not less than 15 is 15, so the median class is $20-30$.

$L=20,\ c.f.=10,\ f=5,\ i=10$

$M_d=L+\frac{\frac{N}{2}-c.f.}{f}\times i=20+\frac{15-10}{5}\times 10 = 30$

The median is 30 years.

(c)

$\bar X=\frac{\Sigma fm}{N}=\frac{850}{30} \approx 28.33$

The average age is 28.33 years (about 28.33).

(d) Two players of age 50-60 (mid-value 55) are added.

New $N=30+2=32$, new $\Sigma fm=850+2\times55=960$

$\text{New mean}=\frac{960}{32}=30$

The new average age is 30 years.

(a) Mode $Z=L+\dfrac{f_1-f_0}{2f_1-f_0-f_2}\times i$, where $L$ is the lower limit of the modal class, $f_1$ its frequency, $f_0$ and $f_2$ the frequencies of the classes before and after it, and $i$ the class size.

(b) The highest frequency is 7, so the modal class is $50-60$.

$L=50,\ f_1=7,\ f_0=6,\ f_2=3,\ i=10$

$Z=L+\frac{f_1-f_0}{2f_1-f_0-f_2}\times i=50+\frac{7-6}{2(7)-6-3}\times 10 = 52$

The modal value is 52 years.

(c) Here $N=\Sigma f=25$.

$\frac{N}{2}=12.5$. The first c.f. not less than 12.5 is 13, so the median class is $40-50$.

$L=40,\ c.f.=7,\ f=6,\ i=10$

$M_d=L+\frac{\frac{N}{2}-c.f.}{f}\times i=40+\frac{12.5-7}{6}\times 10 \approx 49.17$

The median is about 49.17 years.

(d) The median class is 40-50. People below it (20-30, 30-40): $3+4=7$. People above it (50-60, 60-70, 70-80): $7+3+2=12$.

The number of people above the median class (12) is greater than the number below it (7), by 5. Ratio above : below $=12:7$.

(a) $L$ represents the lower limit of the median class.

(b) The table gives cumulative frequencies, so the ordinary frequencies are found by subtraction:

Here $N=\Sigma f=30$.

$\frac{N}{2}=15$. The first c.f. not less than 15 is 17, so the median class is $10-20$.

$L=10,\ c.f.=12,\ f=5,\ i=10$

$M_d=L+\frac{\frac{N}{2}-c.f.}{f}\times i=10+\frac{15-12}{5}\times 10 = 16$

The median is 16 marks.

(c)

$\bar X=\frac{\Sigma fm}{N}=\frac{640}{30} \approx 21.33$

The average mark is about 21.33.

(d) If the students of 0-30 are removed, the remaining students are in 30-40 (2 students) and 40-50 (8 students); $N=10$.

$\text{New mean}=\frac{2\times35+8\times45}{10}=\frac{70+360}{10}=43$

Difference $=43-21.33=21.67$ (exactly $\frac{65}{3}$). The average mark increases by about 21.67.

(a) Two events are independent if the occurrence of one does not affect the occurrence of the other (e.g. drawing two cards with replacement).

(b) F = face card (J, Q, K: 12 cards), N = not a face card (40 cards). With replacement the probabilities stay the same on the second draw.

Start
+-- F (12/52)
|       +-- F (12/52)  ->  FF = 9/169
|       +-- N (40/52)  ->  FN = 30/169
+-- N (40/52)
        +-- F (12/52)  ->  NF = 30/169
        +-- N (40/52)  ->  NN = 100/169

(c) $P(FF)=\frac{12}{52}\times\frac{12}{52}=\frac{3}{13}\times\frac{3}{13}=\frac{9}{169}$

(d) Without replacement: $P(FF)=\frac{12}{52}\times\frac{11}{51}=\frac{11}{221}$

Difference $=\frac{9}{169}-\frac{11}{221}=\frac{9\times17-11\times13}{2873}=\frac{153-143}{2873}=\frac{10}{2873}$

The probability with replacement is greater by $\frac{10}{2873}$ (about 0.0035).

(a) Mode $Z=L+\dfrac{f_1-f_0}{2f_1-f_0-f_2}\times i$, where $L$ = lower limit of the modal class, $f_1$ = its frequency, $f_0$, $f_2$ = frequencies of the preceding and following classes, $i$ = class size.

(b) The highest frequency is 16, so the modal class is $130-140$.

$L=130,\ f_1=16,\ f_0=10,\ f_2=12,\ i=10$

$Z=L+\frac{f_1-f_0}{2f_1-f_0-f_2}\times i=130+\frac{16-10}{2(16)-10-12}\times 10 = 136$

The mode is 136 cm.

(c) Here $N=\Sigma f=50$.

$\frac{3N}{4}=37.5$. The first c.f. not less than 37.5 is 46, so the third quartile class is $140-150$.

$L=140,\ c.f.=34,\ f=12,\ i=10$

$Q_3=L+\frac{\frac{3N}{4}-c.f.}{f}\times i=140+\frac{37.5-34}{12}\times 10 \approx 142.92$

The upper quartile is about 142.92 cm.

(d) The modal class is 130-140. The classes above it are 140-150 ($f=12$, $m=145$) and 150-160 ($f=4$, $m=155$).

$\text{Average height}=\frac{12\times145+4\times155}{12+4}=\frac{1740+620}{16}=\frac{2360}{16}=147.5\text{ cm}$

(a) For events that are not mutually exclusive, $P(G\cup B)=P(G)+P(B)-P(G\cap B)$.

(b) G = girl, B = boy. Total = 50 students; selection is without repetition.

Start
+-- G (27/50)
|       +-- G (26/49)  ->  GG = 351/1225
|       +-- B (23/49)  ->  GB = 621/2450
+-- B (23/50)
        +-- G (27/49)  ->  BG = 621/2450
        +-- B (22/49)  ->  BB = 253/1225

(c) $P(\text{same gender})=P(GG)+P(BB)=\frac{27}{50}\times\frac{26}{49}+\frac{23}{50}\times\frac{22}{49}=\frac{702}{2450}+\frac{506}{2450}=\frac{1208}{2450}=\frac{604}{1225}$

(d) $P(GG)=\frac{702}{2450}=\frac{351}{1225}$. The total probability is 1.

$1-\frac{351}{1225}=\frac{874}{1225}$

The probability of selecting both girls is less than the total probability by $\frac{874}{1225}$ (about 0.713).

(a) Since 34 lies between 30 and 40, the median class is 30-40.

(b) $N=2+3+4+p+6=15+p$. Cumulative frequency before the median class $=2+3+4=9$. For the median class: $L=30$, $f=p$, $i=10$.

$M_d=L+\frac{\frac{N}{2}-c.f.}{f}\times i$

$34=30+\frac{\frac{15+p}{2}-9}{p}\times10$

$4p=10\left(\frac{p-3}{2}\right)=5p-15$

$p=15$

Check: $N=30$, $\frac{N}{2}=15$, $30+\frac{15-9}{15}\times10=34$.

(c) With $p=15$: frequencies 2, 3, 4, 15, 6.

$\bar X=\frac{\Sigma fm}{N}=\frac{950}{30} \approx 31.67$

The mean is about 31.67.

(d) Students below the median class: $2+3+4=9$ out of $N=30$.