A college intramural sports program requires all students to take a fitness test. The time to run one mile is recorded for 200 male students in the program. These times have a normal probability distribution with mean 8 minutes and standard deviation 1 minute. Mike's time for one mile is 9 minutes. What is the probability that a student will take longer than Mike to run a mile?

the probability is P = 0.159

Step-by-step explanation:

The mean of the normal distribution is 8 minutes and the standard deviation is 1 minute.

In a normal distribution the percentages of the sample distribute in the bell in a next way:

in the range of the mean and the mean plus one standard deviation, the percentage is 34.1% of the sample

and for this point in forward, the percentage is 15.9%

This means that 15.9% of the population is in the range of mean + standard deviation in forward.

in this case, the mean + standard deviation is 8 min + 1 min = 9 min (which is the time of mike)

Then the percentage of students that take longer than mike (more than the mean plus one time the standard deviation) is equal to 15.9% of the sample.

then the probability for a randomly selected student is equal to 0.159

Given Information:  

Number of students = n = 200  

Mean = μ = 8 minutes

Standard deviation = σ = 1 minute  

Required Information:  

P(X > 9) = ?  

P(X > 9) = 15.87 %

Explanation:  

Let X time in hours to run one mile is a random variable that has a Normal distribution with mean of 8 minutes and standard deviation of 1 minute.

We want to find out the probability that a student will take longer time than Mike (9 minutes) to run a mile.

P(X > 9) = 1 - P(Z < (x - μ)/σ)

P(X > 9) = 1 - P(Z < (9 - 8)/1)

P(X > 9) = 1 - P(Z < 1)

From the z-table , the z-score corresponding to 1 is 0.84134

P(X > 9) = 1 - 0.84134

P(X > 9) = 0.15866

P(X > 9) = 15.87 %

Therefore, there is 15.87% probability that a student will take longer time than Mike that is 9 minutes.

idk im srry

step-by-step explanation:

answer: true

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