# CIE A level Math -Probability & Statistics 1: 5.5 The normal distribution: problems concerning a variable X : Exam Style Questions Paper 5

The weights of male leopards in a particular region are normally distributed with mean 55 kg and standard deviation 6 kg.

### (a) Question

Find the probability that a randomly chosen male leopard from this region weighs between 46 and 62 kg.

The weights of female leopards in this region are normally distributed with mean 42 kg and standard deviation 3 kg. It is known that 25% of female leopards in the region weigh less than 36 kg.

Ans:

$$P(46< X< 62) = P\left ( \frac{46-55}{6} < Z< \frac{62-55}{6}\right )$$

$$= P \left ( -1.5< Z< \frac{7}{6} \right )$$

$$\left [ =\Phi\left ( \frac{7}{6} \right ) -\left ( 1-\Phi \left ( 1.5 \right ) \right )\right ]$$

= 0.8784  + (0.9332 – 1 )

0.812

### (b) Question

Find the value of 3.

The distributions of the weights of male and female leopards are independent of each other. A male leopard and a female leopard are each chosen at random.

Ans:

z = ±0.674

$$\frac{36-42}{\sigma } = -0.674$$

σ= 8.9 [ 0 ]

### (c) Question

Find the probability that both the weights of these leopards are less than 46 kg.

Ans:

P(male < 46) = 1−their 0.9332 = 0.0668

P(female < 46) = $$P \left ( Z< \frac{46-42}{their 8.90} \right )\left [ =\Phi (0.449) \right ]$$

= 0.6732

P(both) = 0.0668 ×0.6732

0.0450 or 0.0449

### Question

The time in hours that Davin plays on his games machine each day is normally distributed with
mean 3.5 and standard deviation 0.9.
(a) Find the probability that on a randomly chosen day Davin plays on his games machine for more
than 4.2 hours.                                                                                                                                                       [3]

(b) On 90% of days Davin plays on his games machine for more than t hours. Find the value of t.           [3]

(c) Calculate an estimate for the number of days in a year (365 days) on which Davin plays on his
games machine for between 2.8 and 4.2 hours.                                                                                                [3]

Ans

5 (a)  $$P(X\gt 4.2)=P(z\gt \frac{4.2-3.5}{0.9})$$

= P ( z > 0.7778)

1 – 0·7818

0·218

5 (b) z = −1.282

$$\frac{t – 3.5}{0.9}=-1.282$$

t = 2.35

5 (c) P(2.8 < X < 4.2) = 1 – 2 × their 5(a)
≡ 2(1 – their 5(a)) – 1
≡ 2(0·5 – their 5(a))
= 0·5636

Number of days = 365 × 0·5636 = 205·7

So, 205 (days)

Alternative method for question 5(c)

$$P\left( \frac{2.8-3.5}{0.9}\lt z\lt \frac{4.2-3.5}{0.9} \right)$$

= −− Φ(0.7778 ) – ( 1 – Φ0.7778)
= 0·7818 – (1 – 0·7818)
= 0·5636

Number of days = 365 × 0·5636 = 205·7
So, 205 (days)

### Question

The time spent by shoppers in a large shopping centre has a normal distribution with mean 96 minutes
and standard deviation 18 minutes.
(a) Find the probability that a shopper chosen at random spends between 85 and 100 minutes in the
shopping centre.
88% of shoppers spend more than t minutes in the shopping centre.

(b) Find the value of t.

1. $$P((\frac{85-96}{18})<z<(\frac{100-96}{18}))$$
$$-1.175=\frac{t-96}{18}$$