John drives to work each morning, and the trip takes an average of µ = 38 minutes. The distribution of driving times is approximately normal with a standard deviation of σ = 5 minutes. For a randomly selected morning, what is the probability that John’s drive to work will take between 36 and 40 minutes?

the probability that John’s drive to work will take between 36 and 40 minutes is 0.32

Step-by-step explanation:

Since the distribution of driving times is approximately normal, we would apply the formula for normal distribution which is expressed as

z = (x - µ)/σ

Where

x = driving times.

µ = mean time

σ = standard deviation

From the information given,

µ = 38 minutes

σ = 5 minutes

The probability that John’s drive to work will take between 36 and 40 minutes is expressed as

P(36 ≤ x ≤ 40)

For x = 36,

z = (36 - 38)/5 = - 0.4

Looking at the normal distribution table, the probability corresponding to the z score is 0.34

For x = 40,

z = (40 - 38)/5 = 0.4

Looking at the normal distribution table, the probability corresponding to the z score is 0.66

Therefore,

P(36 ≤ x ≤ 40) = 0.66 - 0.34 = 0.32

step-by-step explanation:

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answer: im gonna go with milli

step-by-step explanation: