Question

During the last week of the semester, students at a certain college spend on the average 4.2 hours using the school’s computer terminals with a standard deviation of 1.8 hours. For a random sample of 36 students at that college, find the probabilities that the average time spent using the computer terminals during the last week of the semester is a) At least 4.8 hours (Draw the bell curve) b)

Answer 1

To find the probabilities, we calculate the z-score and use the standard normal distribution table. For at least 4.8 hours, the probability is about 0.6311. For less than 4.8 hours, the probability is about 0.3689.

Step-by-step explanation:

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

a) Probability of spending at least 4.8 hours

Step 1: Calculate the z-score. z = (x - μ) / (σ / √n) = (4.8 - 4.2) / (1.8 / √36) = 0.3333

Step 2: Use the standard normal distribution table to find the probability of getting a z-score greater than 0.3333. The probability is about 0.6311.

b) Probability of spending less than 4.8 hours

Step 1: Calculate the z-score. z = (x - μ) / (σ / √n) = (4.8 - 4.2) / (1.8 / √36) = 0.3333

Step 2: Use the standard normal distribution table to find the probability of getting a z-score less than 0.3333. The probability is about 0.3689.

Answer 2

0.97725

Step-by-step explanation:

Given :

Population mean, μ = 4.2

Population standard deviation, σ = 1.8

Sample size, n = 36

Obtain the Zscore of x = 4.8

Zscore = (x - μ) ÷ (σ / √n)

Zscore = (4.8 - 4.2) ÷ (1.8 / √36)

Zscore = 0.6 ÷ 0.3

Z = 2

Hence,

P(Z ≤ 2)

Using the Z probability calculator :

P(Z ≤ 2) = 0.97725