Question

Can you tell me the next steps to solve this problem?

John drives to work each morning, and the trip takes an average of µ = 36 minutes. The distribution of driving times is approximately normal with a standard deviation of σ = 4 minutes. For a randomly selected morning, what is the probability that John’s drive to work will take between 34 and 41 minutes?

The first step Id take is find the Z-Scores for each XValue:

Z= (34-36)/4 = -2/4 = -.5

Z= (41-36)/4 = 5/4 - 1.25

Answer 1

Given that, mean of µ = 36 minutes and

Standard deviation of σ = 4 minutes

We want to find, P(34 < X < 41)

Z-score for X = 34 is,

Z = (34 - 36)/4 = -2/4 = -0.5

Z-score for X = 41 is,

Z = (41 - 36)/4 = 5/4 = 1.25

So, using the standard normal z-table we find,

P(-0.5 < Z < 1.25)

= P(Z < 1.25) - P(Z < -0.5)

= 0.8944 - 0.3085

= 0.5859

=> P(-0.5 < Z < 1.25) = 0.5859

=> P(34 < X < 41) = 0.5859

Therefore, the probability that John’s drive to work will take between 34 and 41 minutes is 0.5859