Chapter 6

Statistics and Probability

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Past questions and answers

2 questions set from this chapter. Most repeated first.

2081 · Mathematics · 5 marks

a) Find the arithmetic mean of the following data:

Height (m) 10 20 30 40 50 60
No. of plants 2 3 9 15 11 5
b) A fair dice is rolled. What is the probability of getting: i) an even number? ii) a number greater than 3? iii) the number 7?

Height (m)	10	20	30	40	50	60
No. of plants	2	3	9	15	11	5

Answer

a) Arithmetic mean of the given data

Height (m) (x)	No. of plants (f)	(fx)
10	2	20
20	3	60
30	9	270
40	15	600
50	11	550
60	5	300
Total	(\sum f = 45)	(\sum fx = 1800)

[ \text{Arithmetic mean} = \frac{\sum fx}{\sum f} ]

[ = \frac{1800}{45} = 40 ]

So, the arithmetic mean is 40 m.

b) Probability when a fair dice is rolled

A fair dice has sample space:

[ S = {1,2,3,4,5,6}, \quad n(S)=6 ]

i) Probability of getting an even number

Even numbers are (2,4,6), so (n(E)=3).

[ P(\text{even number}) = \frac{3}{6} = \frac{1}{2} ]

ii) Probability of getting a number greater than 3

Numbers greater than 3 are (4,5,6), so (n(E)=3).

[ P(\text{number greater than 3}) = \frac{3}{6} = \frac{1}{2} ]

iii) Probability of getting the number 7

A dice has no number 7, so it is an impossible event.

[ P(\text{getting 7}) = 0 ]

Answers:

i) (\frac{1}{2})
ii) (\frac{1}{2})
iii) (0)

Answer checked by Claude Haiku · Oct 21, 2026

Test yourself · 2 MCQs

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2082 · Mathematics · 2+2+2 marks

A fair coin is tossed 4 times. Let X be number of heads obtained. a) Calculate probability distribution of X. b) Calculate mean and variance of the distribution. c) Justify whether this is a binomial distribution.

Answer

Let (X) = number of heads in 4 tosses of a fair coin.
Here (n=4), (p=\frac12), (q=\frac12).

[ P(X=x)={}^4C_x\left(\frac12\right)^x\left(\frac12\right)^{4-x} =\frac{{}^4C_x}{16} ]

a) Probability distribution of (X)

(X=x)	Calculation	(P(X=x))
0	(\frac{{}^4C_0}{16}=\frac{1}{16})	(\frac{1}{16})
1	(\frac{{}^4C_1}{16}=\frac{4}{16})	(\frac14)
2	(\frac{{}^4C_2}{16}=\frac{6}{16})	(\frac38)
3	(\frac{{}^4C_3}{16}=\frac{4}{16})	(\frac14)
4	(\frac{{}^4C_4}{16}=\frac{1}{16})	(\frac{1}{16})

So,

[ P(X=0)=\frac1{16},\quad P(X=1)=\frac14,\quad P(X=2)=\frac38,\quad P(X=3)=\frac14,\quad P(X=4)=\frac1{16} ]

b) Mean and variance

For binomial distribution,

[ \text{Mean}=np=4\times \frac12=2 ]

[ \text{Variance}=npq=4\times \frac12\times \frac12=1 ]

Hence,

[ \boxed{\text{Mean}=2,\qquad \text{Variance}=1} ]

c) Justification

Yes, (X) follows a binomial distribution because:

The number of trials is fixed: (n=4).
Each trial has only two outcomes: head or tail.
The trials are independent.
The probability of success, i.e. getting a head, is constant: (p=\frac12).

Therefore,

[ \boxed{X\sim B\left(4,\frac12\right)} ]

So the given experiment is a binomial distribution.