# Exam-style Questions: The Normal Distribution

**1.** The time taken for a paperboy to deliver his papers is normally distributed with mean 52 minutes and standard deviation 6.5 minutes.

**Find the probability on any given day the paperboy takes:**

**a)** longer than 1 hour to deliver

**(3 marks)**

**b)** less than 45 minutes

**(3 marks)**

**c)** between 48 and 56 minutes

**(3 marks)**

**(Marks available: 9)**

**Answer outline and marking scheme for question: 1**

**1.** X ~ N(52, 6.52)

**a)** P(X > 60) = P(Z > 60 - 52) add bell on ans sheet 3 A1

6.5

= P(Z > 1.23) = 1 - F(1.23) A1

= 1 - 0.8907 = 0.1093 A1

**(3 marks)**

**b)** P(X

6.5

= P(Z = 1 - F(1.08)

= 1 - 0.8599 = 0.1401 A1

**(3 marks)**

**c)** P(48

6.5 6.5

= P(-0.62

= F(0.62) - [1-F(0.62)] A1

= 2F(0.62) - 1 = 2 ' 0.7324 - 1

= 0.4648 A1

**(3 marks)**

**(Marks available: 9)**

**2.** An Olympic high jumper jumps distances which are normally distributed with mean, μ = 2.45m and variance 0.49m.

**a)** find the probability the jumper manages a height over 2.35m

**(2 marks)**

**b)** What distance is exceeded by 5% of his jumps?

**(4 marks)**

**(Marks available: 6)**

**Answer outline and marking scheme for question: 2**

X ~ N(2.45,0.49) we are given the variance so s = v0.49= 0.7

**a)** P(X > 2.35) = P(Z > 2.35 - 2.45) add bell on sheet 3

0.7

= P(Z > -0.14) = F(0.14) = 0.5557

**(2 marks)**

**b)** we need the distance d where

P(X > d) = 0.05

P(Z > d - 2.45) = 0.05 sheet 3 for bell

0.7

1 - F(d - 2.45) = 0.05 hence F(d - 2.45) = 0.95

0.7 0.7

so d - 2.45 = 1.645 hence d = 3.6m ( a very high jump!)

0.7

**(4 marks)**

**(Marks available: 6)**