The time needed to complete a final examination in a particular college course is normally distributed with a mean of 83 minutes and a standard deviation of 13 minutes. Answer the following questions.

What is the probability of completing the exam in one hour or less (to 4 decimals)?

What is the probability that a student will complete the exam in more than 60 minutes but less than 75 minutes (to 4 decimals)?

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 (to the nearest whole number)?

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To calculate the probabilities, we need to use the z-score formula and the standard normal distribution table.

1. Probability of completing the exam in one hour or less (60 minutes):

First, we need to calculate the z-score using the formula:
z = (x - μ) / σ

where x is the value we want to find the probability for (60 minutes), μ is the mean (83 minutes), and σ is the standard deviation (13 minutes).

z = (60 - 83) / 13 = -23 / 13 ≈ -1.7692

Now, we look up the z-score in the standard normal distribution table. The table gives us the probability to the left of the z-score. In this case, we want the probability to the right since we want the probability of completing the exam in one hour or less.

From the table, we find that the probability to the left of -1.7692 is 0.0384 (rounded to four decimal places).

To find the probability to the right, we subtract the probability to the left from 1:
Probability = 1 - 0.0384 = 0.9616 (rounded to four decimal places)

Therefore, the probability of completing the exam in one hour or less is approximately 0.9616.

2. Probability of completing the exam in more than 60 minutes but less than 75 minutes:

First, we calculate the z-scores for both values:
z1 = (60 - 83) / 13 = -23 / 13 ≈ -1.7692
z2 = (75 - 83) / 13 = -8 / 13 ≈ -0.6154

Now, we can look up the probabilities corresponding to these z-scores in the standard normal distribution table.

The probability to the left of -1.7692 is 0.0384 (from the previous calculation).

The probability to the left of -0.6154 is 0.2681 (rounded to four decimal places).

To find the probability between these two values, we subtract the second probability from the first:
Probability = 0.0384 - 0.2681 = -0.2297 (rounded to four decimal places)

However, probabilities cannot be negative, so we take the absolute value:
Probability = | -0.2297 | = 0.2297 (rounded to four decimal places)

Therefore, the probability of completing the exam in more than 60 minutes but less than 75 minutes is approximately 0.2297.

3. Number of students unable to complete the exam in the allotted time:

Since the examination period is 90 minutes, any student taking longer than 90 minutes will not be able to complete the exam in the allotted time.

We need to find the probability that a student takes longer than 90 minutes. To do this, we convert 90 minutes into a z-score.

z = (90 - 83) / 13 = 7 / 13 ≈ 0.5385

Looking up this z-score in the standard normal distribution table, we find the probability to the left of 0.5385 is 0.7032 (rounded to four decimal places).

The probability of a student taking longer than 90 minutes is 1 - 0.7032 = 0.2968 (rounded to four decimal places).

Now, we can calculate the number of students who will be unable to complete the exam:
Number of students = Probability of taking longer than 90 minutes * Number of students (60)
Number of students = 0.2968 * 60 ≈ 17.808 ≈ 18 (rounded to the nearest whole number)

Therefore, we expect approximately 18 students to be unable to complete the exam in the allotted time.

To answer these questions, we need to use the properties of the normal distribution. Specifically, we'll use the Z-score formula:

Z = (X - μ) / σ

Where:
Z is the standard score,
X is the value we're interested in,
μ is the mean of the distribution, and
σ is the standard deviation of the distribution.

Now, let's answer each question step by step:

1. Probability of completing the exam in one hour or less (to 4 decimals):

To convert one hour (60 minutes) into a Z-score, we use the formula:

Z = (60 - 83) / 13 = -23 / 13 ≈ -1.7692

By referring to a Z-table or using statistical software, we can find that the probability of a Z-score being less than -1.7692 is approximately 0.0384.

Therefore, the probability of completing the exam in one hour or less is 0.0384 (to 4 decimals).

2. Probability of completing the exam in more than 60 minutes but less than 75 minutes (to 4 decimals):

First, we find the Z-scores for both 60 and 75 minutes:

Z1 = (60 - 83) / 13 ≈ -1.7692
Z2 = (75 - 83) / 13 ≈ -0.6154

Using the Z-table or software, we find the probability of a Z-score falling between -1.7692 and -0.6154 is approximately 0.1681.

Therefore, the probability of completing the exam in more than 60 minutes but less than 75 minutes is 0.1681 (to 4 decimals).

3. Number of students who are unable to complete the exam in the allotted time (to the nearest whole number):

We'll use the standardized Z-score formula again, but this time we'll find the Z-score for 90 minutes since the exam period is 90 minutes:

Z = (90 - 83) / 13 ≈ 0.5385

Now, we want to find the probability of completing the exam in more than 90 minutes, which is equivalent to finding the area to the right of the Z-score of 0.5385 on the standard normal distribution.

By using the Z-table or software, we find that the probability is approximately 0.2946.

To find the number of students unable to complete the exam in the allotted time, we multiply the probability by the total number of students (60):

Number of students = Probability * Total number of students
Number of students = 0.2946 * 60 ≈ 17.678

Rounding to the nearest whole number, we expect around 18 students to be unable to complete the exam in the allotted time.