Question

The mean one-way commute to work in Chowchilla is 7 minutes. The standard deviation is 2.4 minutes, and the population is normally distributed. What is the probability of randomly selecting one commute time and finding that: a). P (x < 2 mins) _____________________________ b). P (2 < x < 11 mins) _____________________________ c). P (x < 11 mins) ________________________________ d). P (2 < x < 5 mins) _______________________________ e). P (x > 5 mins)

Answer 1

The answer is below

Explanation:

Given that:

The mean (μ) one-way commute to work in Chowchilla is 7 minutes. The standard deviation (σ) is 2.4 minutes.

The z score is used to determine by how many standard deviations the raw score is above or below the mean. It is given by:

`z=(x-\mu)/(\sigma)`

a) For x < 2:

`z=(x-\mu)/(\sigma)=(2-7)/(2.4) =-2.08`

From normal distribution table, P(x < 2) = P(z < -2.08) = 0.0188 = 1.88%

b) For x = 2:

`z=(x-\mu)/(\sigma)=(2-7)/(2.4) =-2.08`

For x = 11:

`z=(x-\mu)/(\sigma)=(11-7)/(2.4) =1.67`

From normal distribution table, P(2 < x < 11) = P(-2.08 < z < 1.67 ) = P(z < 1.67) - P(z < -2.08) = 0.9525 - 0.0188 = 0.9337

c) For x = 11:

`z=(x-\mu)/(\sigma)=(11-7)/(2.4) =1.67`

From normal distribution table, P(x < 11) = P(z < 1.67) = 0.9525

d) For x = 2:

`z=(x-\mu)/(\sigma)=(2-7)/(2.4) =-2.08`

For x = 5:

`z=(x-\mu)/(\sigma)=(5-7)/(2.4) =-0.83`

From normal distribution table, P(2 < x < 5) = P(-2.08 < z < -0.83 ) = P(z < -0.83) - P(z < -2.08) = 0.2033- 0.0188 = 0.1845

e) For x = 5:

`z=(x-\mu)/(\sigma)=(5-7)/(2.4) =-0.83`

From normal distribution table, P(x < 5) = P(z < -0.83) = 0.2033