Question

In a study of sleep quality and academic performance, the daily sleep duration among college students was approximately Normally distributed with means = 7.13 hours and standard deviation o = 1.67 hours. You plan to take an SRS of size * = 60 and compute the average total sleep time. What is the probability that your average will be below 6.9 hours? Provide your answer to three decimal places. P(X < 6,9)

Answer 1

The probability that the average sleep time will be below 6.9 hours is 0.445, or 44.5% (rounded to three decimal places).

Step-by-step explanation:

We can solve this problem by using the z-score formula. The formula to calculate z-score is:

z = (x - μ) / σ

Where:

• z is the z-score
• x is the value we want to find the probability for
• μ is the mean
• σ is the standard deviation

In this case, we want to find the probability that the average sleep time is below 6.9 hours, which means we need to find P(X < 6.9).

Using the z-score formula, we can calculate the z-score:

z = (6.9 - 7.13) / 1.67

z = -0.138

Now, we can use a z-score table or a calculator to find the probability associated with this z-score.

Using a z-score table or calculator, we find that the probability associated with a z-score of -0.138 is approximately 0.445.

Therefore, the probability that the average sleep time will be below 6.9 hours is 0.445, or 44.5% (rounded to three decimal places).

Answer 2

To find the probability that the average sleep duration will be below 6.9 hours, we need to standardize the average using the z-score formula.

Step-by-step explanation:

To find the probability that the average sleep duration will be below 6.9 hours, we need to standardize the average using the z-score formula.

The formula for calculating the z-score is: z = (X - μ) / (σ / √n) where X is the average, μ is the mean, σ is the standard deviation, and n is the sample size.

Plugging in the values, we have: z = (6.9 - 7.13) / (1.67 / √60)

Calculating the z-score, we get z ≈ -1.250.

We can then look up the corresponding probability in the standard normal distribution table or use a calculator to find that the probability is approximately 0.106.