"Statistics and Probability" — Practice

Question 1

Find the mean, median and mode of $4, 7, 7, 9, 12, 15, 15, 15$.

Accepted Answer

The sum is 84 over 8 values, so the mean is 10.5. With eight values the median is the average of the fourth and fifth, that is $(9+12)/2 = 10.5$. The mode is 15, occurring three times. Here the mean and median coincide while the mode is well above both, which signals that the repeated large value is pulling the mode away from the centre rather than the distribution being genuinely skewed.

Question 2

For classes $0$–$10$, $10$–$20$, $20$–$30$, $30$–$40$, $40$–$50$ with frequencies $4, 10, 18, 12, 6$, find the median.

Accepted Answer

The total frequency is 50, so $n/2 = 25$. The cumulative frequencies are 4, 14, 32, 44, 50, and 25 first falls inside the class $20$–$30$, whose lower boundary is 20, width 10, preceding cumulative frequency 14 and frequency 18. The median is $20 + [(25-14)/18]	imes 10 = 20 + 110/18$, which is about 26.11. Note that the median class is found from the cumulative column, not from the largest frequency, which locates the mode instead.

Question 3

Find the variance and standard deviation of $6, 8, 10, 12, 14$.

Accepted Answer

The mean is 10. Using the shortcut, the sum of squares is $36+64+100+144+196 = 540$, so the variance is $540/5 - 100 = 108 - 100 = 8$. The standard deviation is $\sqrt8 = 2\sqrt2$, about 2.83. Checking directly, the deviations are $-4, -2, 0, 2, 4$ with squares summing to 40, and $40/5 = 8$ as required.

Question 4

The mean of five observations is 4.4 and their variance is 8.24. Three of them are 1, 2 and 6. Find the other two.

Accepted Answer

The sum is $5 	imes 4.4 = 22$, so the two unknowns total $22 - 9 = 13$. From the variance, the sum of squares is $5(8.24 + 4.4^2) = 5(8.24 + 19.36) = 138$. The known squares total $1 + 4 + 36 = 41$, so the two unknown squares total 97. From $a+b = 13$ and $a^2+b^2 = 97$ we get $2ab = 169 - 97 = 72$, so $ab = 36$. The two numbers are the roots of $t^2 - 13t + 36 = 0$, which are 4 and 9.

Question 5

The variance of a dataset is 20. Each value is multiplied by 3 and then increased by 5. Find the new variance.

Accepted Answer

Multiplying every value by 3 multiplies the variance by $3^2 = 9$, giving 180. Adding 5 afterwards shifts the whole dataset without stretching it, so the variance is unchanged by that step. The new variance is 180. The new standard deviation is $\sqrt{180}$, which is 3 times the original $\sqrt{20}$, confirming that standard deviation scales by $k$ while variance scales by $k$ squared.

Question 6

Two fair dice are thrown. Find the probability that the sum is 8.

Accepted Answer

There are 36 equally likely outcomes. The sum is 8 for $(2,6)$, $(3,5)$, $(4,4)$, $(5,3)$ and $(6,2)$, which is five outcomes, so the probability is $5/36$. Note that $(2,6)$ and $(6,2)$ are counted separately because the dice are distinguishable, while $(4,4)$ occurs only once. Treating the sums as equally likely instead, and answering $1/11$, is the classic error here.

Question 7

$P(A) = 0.4$, $P(B) = 0.5$ and $P(A\cap B) = 0.2$. Are $A$ and $B$ independent? Find $P(A\cup B)$ and $P(A\mid B)$.

Accepted Answer

Test independence against the definition: $P(A)P(B) = 0.4 	imes 0.5 = 0.2$, which equals $P(A\cap B)$, so the events are independent. The union is $0.4 + 0.5 - 0.2 = 0.7$. The conditional probability is $0.2/0.5 = 0.4$, which equals $P(A)$, confirming independence from the other direction: knowing $B$ occurred does not change the chance of $A$.

Question 8

$A$ and $B$ are mutually exclusive with $P(A) = 0.3$ and $P(B) = 0.4$. Find $P(A\mid B)$ and explain why they cannot be independent.

Accepted Answer

Mutually exclusive means $P(A\cap B) = 0$, so $P(A\mid B) = 0/0.4 = 0$. Independence would require $P(A\mid B) = P(A) = 0.3$, which it is not. More generally, independence would need $P(A\cap B) = P(A)P(B) = 0.12$, but exclusivity forces it to be 0, and $0.12$ is not 0. Since both probabilities are non-zero, the two conditions are incompatible: exclusivity is the strongest possible dependence.

Question 9

Three machines produce 25, 35 and 40 per cent of a factory's output, with defect rates 5, 4 and 2 per cent. A defective item is found. What is the probability it came from the third machine?

Accepted Answer

The total probability of a defect is $0.25(0.05) + 0.35(0.04) + 0.40(0.02) = 0.0125 + 0.014 + 0.008 = 0.0345$. Bayes then gives $P(M_3\mid D) = 0.008/0.0345 = 16/69$, about 0.232. Notice that the third machine makes the most items but has the lowest defect rate, and the two effects nearly cancel, leaving it responsible for under a quarter of the defects despite producing 40 per cent of the output.

Question 10

A man speaks the truth 4 times out of 5. He throws a die and reports a six. What is the probability that it really was a six?

Accepted Answer

The prior for a six is $1/6$, and he reports a six truthfully with probability $4/5$. If it was not a six, which has prior $5/6$, he lies with probability $1/5$ and reports a six. So the total probability of the report is $(1/6)(4/5) + (5/6)(1/5) = 4/30 + 5/30 = 9/30$. Bayes gives $(4/30)/(9/30) = 4/9$, about 0.444. Under a half, even though he is usually truthful, because a six was unlikely to begin with: that is the prior doing its work.

Question 11

A fair coin is tossed 6 times. Find the probability of exactly 4 heads.

Accepted Answer

The trials are independent with constant $p = 1/2$, so the binomial applies: $P(X=4) = \binom{6}{4}(1/2)^4(1/2)^2 = 15/64$, about 0.234. Since $p = q = 1/2$, every sequence of six tosses is equally likely and the answer is simply the number of ways to choose which four tosses are heads, out of the 64 possible sequences.

Question 12

A binomial distribution has mean 4 and variance 3. Find $n$, $p$ and $P(X=1)$.

Accepted Answer

From $np = 4$ and $npq = 3$, dividing gives $q = 3/4$, so $p = 1/4$ and $n = 4/p = 16$. Then $P(X=1) = \binom{16}{1}(1/4)(3/4)^{15} = 4(3/4)^{15}$, which is $3^{15}/4^{14}$, about 0.0534. The variance being less than the mean was guaranteed before any calculation, since $q$ is always below 1, and a question giving a variance above the mean would have no binomial solution at all.

Practice — "Statistics and Probability"

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