Questions: Suppose the monthly time students at a particular university spend studying during a semester is approximately normally distributed with a mean of 75 hours and a standard deviation of 11 hours. Complete parts a through c. a. What is the probability that an individual student randomly selected from the population will study between 70 and 80 hours? P(70 ≤ x̄ ≤ 80)= (Round to four decimal places as needed.)

Suppose the monthly time students at a particular university spend studying during a semester is approximately normally distributed with a mean of 75 hours and a standard deviation of 11 hours. Complete parts a through c.
a. What is the probability that an individual student randomly selected from the population will study between 70 and 80 hours?
P(70 ≤ x̄ ≤ 80)= (Round to four decimal places as needed.)

Transcript text: Suppose the monthly time students at a particular university spend studying during a semester is approximately normally distributed with a mean of 75 hours and a standard deviation of 11 hours. Complete parts a through $c$. a. What is the probability that an individual student randomly selected from the population will study between 70 and 80 hours? $\mathrm{P}(70 \leq \bar{x} \leq 80)=$ $\square$ (Round to four decimal places as needed.)

Solution

Solution Steps

Step 1: Define the Problem

We need to find the probability that an individual student randomly selected from the population will study between 70 and 80 hours. The study time is normally distributed with a mean ($ \mu $) of 75 hours and a standard deviation ($ \sigma $) of 11 hours.

Step 2: Calculate Z-scores

To find the probability, we first calculate the Z-scores for the lower and upper bounds of the range:

For the lower bound ($ x = 70 $): \[ Z_{start} = \frac{70 - 75}{11} = -0.4545 \]
For the upper bound ($ x = 80 $): \[ Z_{end} = \frac{80 - 75}{11} = 0.4545 \]

Step 3: Calculate the Probability

Using the Z-scores, we can find the probability that an individual student studies between 70 and 80 hours: \[ P(70 \leq X \leq 80) = \Phi(Z_{end}) - \Phi(Z_{start}) = \Phi(0.4545) - \Phi(-0.4545) \] From the calculations, we find: \[ P(70 \leq X \leq 80) = 0.3506 \]