The time needed to complete a final examination in a particular college course is normally distributed with a mean of 79 minutes and a standard deviation of 8 minutes. Answer the following questions. a. What is the probability of completing the exam in one hour or less (to 4 decimals)? b. What is the probability that a student will complete the exam in more than minutes but less than minutes (to 4 decimals)? c. Assume that the class has students and that the examination period is minutes in length. How many students do you expect will be unable to complete the exam in the allotted time (to nearest whole number)?

a) P(x ≤ 60) = 0.0087

b) P(60 < x < 75) = 0.3009

c) About 6 students will be unable to finish the exam in the allotted time

Step-by-step explanation:

This is a normal distribution problem with

Mean = μ = 79 minutes

Standard deviation = σ = 8 minutes.

a) The probability of completing the exam in one hour or less. P(x ≤ 60)

We first standardize/normalize 60 minutes

The standardized score for any value is the value minus the mean then divided by the standard deviation.

z = (x - μ)/σ = (60 - 79)/8 = - 2.38

To determine the probability of completing the exam in one hour or less.

P(x ≤ 60) = P(z ≤ -2.38)

We'll use data from the normal probability table for these probabilities

P(x ≤ 60) = P(z ≤ -2.38) = 0.00866 = 0.0087 to 4d.p

b) The probability that a student will complete the exam in more than 60 minutes but less than 75 minutes. P(60 < x < 75)

We first standardize/normalize 60 and 75 minutes

For 60 minutes

z = (x - μ)/σ = (60 - 79)/8 = - 2.38

For 75 minutes

z = (x - μ)/σ = (75 - 79)/8 = - 0.50

To determine the probability that a student will complete the exam in more than 60 minutes but less than 75 minutes.

P(60 < x < 75) = P(-2.38 < z < -0.50)

We'll use data from the normal probability table for these probabilities

P(60 < x < 75) = P(-2.38 < z < -0.50)

= P(z < -0.50) - P(z < -2.38)

= 0.30954 - 0.00866

= 0.30088 = 0.3009

c. Assume that the class has 60 students and that the examination period is 90 minutes in length. How many students do you expect will be unable to complete the exam in the allotted time?

Probability of finishing in 90 minutes or less = P(x ≤ 90)

We first standardize/normalize 90 minutes

z = (x - μ)/σ = (90 - 79)/8 = 1.38

To determine the probability of completing the exam in 90 minutes or less.

P(x ≤ 90) = P(z ≤ 1.38)

We'll use data from the normal probability table for these probabilities

P(x ≤ 90) = P(z ≤ 1.38) = 0.91621

This means 91.621% of the class will finish just in time.

91.621% × 60 = 54.97 ≈ 54 (we can't round the number up because it represents number of students that finish in time and rounding a 0.97 to a 1 makes the analysis wrong)

About 54 students will finish in time.

Meaning about 6 students will be unable to finish the exam in the allowed time.

Hope this Helps!!!

the least common denominatore of 7 and 2 is 9

step-by-step explanation:

7+2=9