a. The area to the left of -1.42 is 0.0808.
b. The area to the left of -0.58 is 0.7190 and the area to the left of -2.25 is 0.0122.
c. The area to the right of -1.42 is 1 - 0.0808 = 0.9192.
a. What is the probability of completing the exam in one hour or less (to 4 decimals)?
To calculate the probability of completing the exam in one hour or less, we need to find the area under the normal curve to the left of 60 minutes. We can use a z-score to standardize the value of 60 minutes:
z = (60 - μ) / σ = (60 - 77) / 12 = -1.42
Using a standard normal table, we can find that the area to the left of -1.42 is 0.0808. This means that there is a 0.0808 probability, or 8.08%, that a student will complete the exam in one hour or less.
b. What is the probability that a student will complete the exam in more than 50 minutes but less than 70 minutes (to 4 decimals)?
To calculate the probability of completing the exam in more than 50 minutes but less than 70 minutes, we need to find the area under the normal curve between 50 and 70 minutes. We can use z-scores to standardize the values of 50 and 70 minutes:
z_50 = (50 - μ) / σ = (50 - 77) / 12 = -2.25
z_70 = (70 - μ) / σ = (70 - 77) / 12 = -0.58
Using a standard normal table, we can find the area to the left of -0.58 is 0.7190 and the area to the left of -2.25 is 0.0122.
The area between -0.58 and -2.25 is the difference between these two areas, or 0.7190 - 0.0122 = 0.7068.
This means that there is a 0.7068 probability, or 70.68%, that a student will complete the exam in more than 50 minutes but less than 70 minutes.
c. Assume that the class has 100 students and that the examination period is 60 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)?
To calculate the number of students expected to be unable to complete the exam in the allotted time, we need to find the area under the normal curve to the right of 60 minutes. We can use a z-score to standardize the value of 60 minutes:
z = (60 - μ) / σ = (60 - 77) / 12 = -1.42
Using a standard normal table, we can find that the area to the right of -1.42 is 1 - 0.0808 = 0.9192. This means that there is a 0.9192 probability, or 91.92%, that a student will be unable to complete the exam in the allotted time.
Therefore, we expect 91.92% of the class, or 91.92 students, to be unable to complete the exam in the allotted time.
Rounding to the nearest whole number, we expect 92 students to be unable to complete the exam in the allotted time.