Question

Suppose the speeds that people drive down the Pat Bay Highway are normally distributed with a mean of 90 km/hr and a standard deviation of 7 km/hr. Answer each of the following.

(a) What proportion of drivers are travelling between 80 and 120 km/hr? To answer this question, convert the probability question that is being asked to a probability question regarding Z

How do you get 2.57 for Z2

Answer 1

92.3%

Explanation:

If a continuous random variable X is normally distributed with mean μ and variance σ², it is written as:

`\boxed{X \sim\text{N}(\mu,\sigma^2)}`

Given:

• Mean μ = 90
• Standard deviation σ = 7

Therefore, if the speeds that people drive down the Pat Bay Highway are normally distributed:

`\boxed{X \sim\text{N}(90,7^2)}`

where X is the speed in km/h.

f we want to find what proportion of drivers travel between 80 and 120 km/h, we need to find P(80 ≤ X ≤ 120).

`\implies \sf P(80 \leq X \leq 120)=P(X \leq 120)-P(X \leq 80)`

Converting to the Z distribution:

`\boxed{\textsf{If }\: X \sim\textsf{N}(\mu,\sigma^2)\:\textsf{ then }\: (X-\mu)/(\sigma)=Z, \quad \textsf{where }\: Z \sim \textsf{N}(0,1)}`

Transform X to Z:

`\sf P(X \leq 120)=P\left(Z \leq (120-90)/(7)\right)=P\left(Z \leq (30)/(7)\right)`

`\sf P(X \leq 80)=P\left(Z \leq (80-90)/(7)\right)=P\left(Z \leq -(10)/(7)\right)`

Therefore:

`\begin{aligned}\implies \sf P(80 \leq X \leq 120)&=\sf P\left( -(10)/(7) \leq Z \leq (30)/(7)\right)\\\\&=\sf P\left(Z \leq (30)/(7)\right)-P\left(Z \leq -(10)/(7)\right)\\\\&=\sf 0.999990892...-0.0765637255...\\\\&= \sf 0.9234271...\\\\&=\sf 92.34271...\%\end{aligned}`

Therefore, the proportion of drivers travelling between 80 and 120 km/h is 92.3% (nearest tenth).