# CIE A level Math -Probability & Statistics 1: 5.5 The normal distribution: approximation to the binomial distribution : Exam Style Questions Paper 5

### Question

The lengths of female snakes of a particular species are normally distributed with mean 54 cm and
standard deviation 6.1 cm.
(a) Find the probability that a randomly chosen female snake of this species has length between
50 cm and 60 cm.                                                                                                                                                        

The lengths of male snakes of this species also have a normal distribution. A scientist measures the
lengths of a random sample of 200 male snakes of this species. He finds that 32 have lengths less than
45 cm and 17 have lengths more than 56 cm.
(b) Find estimates for the mean and standard deviation of the lengths of male snakes of this species.       

Ans

6 (a) $$P\left ( \frac{50-54}{6.1}< z< \frac{60-54}{6.1} \right )=P(-0.6557< Z< 0.9836)$$

Both values correct

Φ (0.9836) – Φ (–0.6557) = Φ (0.9836) + Φ (0.6557) – 1
= 0.8375 + 0.7441 – 1
(Correct area)

0.582

6 (b) $$\frac{45-\mu }{\sigma }=-0.994$$

$$\frac{56-\mu }{\sigma }=-1.372$$

One appropriate standardisation equation with  µ, σ ,z-value (not probability) and 45 or 56.
11 = 2.366σ
(M1 for correct algebraic elimination of µ or σ from their two simultaneous equations to form an equation in one variable)
σ = 4.65, μ = 49.6

### Question

In Greenton, 70% of the adults own a car. A random sample of 8 adults from Greenton is chosen.
(a) Find the probability that the number of adults in this sample who own a car is less than 6.
A random sample of 120 adults from Greenton is now chosen.
(b) Use an approximation to find the probability that more than 75 of them own a car.

Ans:

(a) 1 – P(6,7,8)
$$=1-(^8C_6 0.7^6 0.3^2 + ^8C_7 0.3^1 +0.7^8)$$
= 1 – 0.55177
= 0.448
Alternative method for question 5(a)
P(0,1,2,3,4,5)
$$=0.3^8 + ^8C_10.7^1 0.3^7 + ^8C_2 0.7^2 0.3^6 +^8C_30.7^3 0.3^5 + ^8C_4 0.7^4 0.3^4 + ^8C_5 0.7^50.3^3$$
=0.448
(b) Mean = $$120 \times 0.7 = 84$$
Var =$$120 \times 0.7 \times 0.3 = 25.2$$
P( more than 75 ) = P$$(z>\frac{75.5-84}{\sqrt{25.2}}$$
$$P(z>-1.693)$$
= 0.955

Scroll to Top