Question

Estimated demand for cars is normally distributed with a standard deviation of 20 and a mean of 120. What is the probability that:

(i)More than 160 cars will be needed. (4 marks)

(ii)Less than 90 cars will be needed. (4 marks)

(iii)More than 100 but less than 130 cars will be needed.(4 marks)

Answer 1

Mean μ = 120

SD σ = 20

(i)

$P(X>160) = 1 -P(X<130) \\ = 1 -P(\frac{X-μ}{σ} < \frac{160-120}{20} ) \\ = 1 -P(Z<2) \\ = 1 -0.9772 \\ = 0.0228$

(ii)

$P(X<90) = P(\frac{X-μ}{σ} < \frac{90-120}{20}) \\ = P(Z<-1.5) \\ = 0.0668$

(iii)

$P(100<X<130) = P(\frac{100-120}{20}<\frac{X-μ}{σ}< \frac{130-120}{20}) \\ = P(-1<Z<0.5) \\ = P(Z<0.5) -P(Z<-1) \\ = 0.6914 -0.1586 \\ = 0.5328$