Over the past 10 years, the local school district has

measured physical fitness for all high school freshmen.
During that time, the average score on a treadmill
endurance task has been μ � 19.8 minutes with a
standard deviation of � � 7.2 minutes. Assuming that
the distribution is approximately normal, find each of
the following probabilities.
a. What is the probability of randomly selecting
a student with a treadmill time greater than
25 minutes? In symbols, p(X � 25) � ?
b. What is the probability of randomly selecting a
student with a time greater than 30 minutes? In
symbols, p(X � 30) � ?
c. If the school required a minimum time of 10 minutes
for students to pass the physical education course,
what proportion of the freshmen would fail?

To find the probabilities in this problem, we need to use the concept of z-scores and the standard normal distribution.

a. Probability of randomly selecting a student with a treadmill time greater than 25 minutes (p(X > 25)):

First, we need to calculate the z-score for the value 25 using the formula:
z = (x - μ) / σ
where x is the value (25), μ is the mean (19.8), and σ is the standard deviation (7.2).

Substituting the values into the formula:
z = (25 - 19.8) / 7.2
z ≈ 0.722

Now we need to find the probability corresponding to z = 0.722 from the standard normal distribution table or using a calculator. The probability of selecting a student with a treadmill time greater than 25 minutes is equal to 1 minus the probability of selecting a student with a treadmill time less than or equal to 25 minutes:
p(X > 25) ≈ 1 - P(Z ≤ 0.722)

b. Probability of randomly selecting a student with a time greater than 30 minutes (p(X > 30)):

Similarly, we calculate the z-score for the value 30:
z = (30 - 19.8) / 7.2
z ≈ 1.361

Now, we find the probability corresponding to z = 1.361 from the standard normal distribution table or using a calculator:
p(X > 30) ≈ 1 - P(Z ≤ 1.361)

c. Proportion of freshmen who would fail (treadmill time less than 10 minutes):

To find this proportion, we calculate the z-score for the value 10:
z = (10 - 19.8) / 7.2
z ≈ -1.354

Now, we find the probability corresponding to z = -1.354 from the standard normal distribution table or using a calculator:
p(X < 10) = P(Z ≤ -1.354).

This probability represents the proportion of freshmen who would fail the physical education course.

Remember to consult the standard normal distribution table or use a calculator with a normal distribution function to find the probabilities associated with each z-score.

To answer these questions, we will use the z-score formula:

z = (X - μ) / σ

where X is the value we are interested in, μ is the mean, and σ is the standard deviation.

a. To find the probability of randomly selecting a student with a treadmill time greater than 25 minutes (p(X > 25)), we need to find the area under the curve to the right of the z-score for 25.

First, let's calculate the z-score for 25:

z = (25 - 19.8) / 7.2 = 0.8333

Using a standard normal distribution table or a calculator, we can find that the area to the right of the z-score 0.8333 is approximately 0.2023.

Therefore, p(X > 25) = 0.2023.

b. To find the probability of randomly selecting a student with a time greater than 30 minutes (p(X > 30)), we follow the same process as in part a.

The z-score for 30 is:

z = (30 - 19.8) / 7.2 = 1.4167

From the standard normal distribution table or using a calculator, we find that the area to the right of the z-score 1.4167 is approximately 0.0793.

Therefore, p(X > 30) = 0.0793.

c. To find the proportion of freshmen who would fail the physical education course with a minimum time requirement of 10 minutes, we need to find the proportion of students with a time less than 10 minutes (p(X < 10)).

The z-score for 10 is:

z = (10 - 19.8) / 7.2 = -1.3542

Based on the z-score and using the standard normal distribution table or a calculator, we find that the area to the left of the z-score -1.3542 is approximately 0.0886.

Therefore, the proportion of freshmen who would fail the course is approximately 0.0886 or 8.86%.

Note: In all these calculations, we are approximating the normal distribution to make the calculations easier.

a. Z = (score-mean)/SD

Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion/probability related to the Z score.

b, c. Same process.