Explanation:
We can use the standard normal distribution to calculate probabilities for a normal distribution with mean 33 and standard deviation 4. We just need to standardize the intervals using the formula:
z = (x - mu) / sigma
where x is the specific value in the interval, mu is the mean, sigma is the standard deviation, and z is the corresponding z-score.
a. Between 29 and 37:
z1 = (29 - 33) / 4 = -1
z2 = (37 - 33) / 4 = 1
Using a standard normal distribution table, the cumulative probability of z being between -1 and 1 is approximately 0.6827.
So the probability that a randomly selected x-value from the distribution is between 29 and 37 is approximately 0.6827.
b. Between 33 and 45:
z1 = (33 - 33) / 4 = 0
z2 = (45 - 33) / 4 = 3
The cumulative probability of z being between 0 and 3 is approximately 0.4987.
So the probability that a randomly selected x-value from the distribution is between 33 and 45 is approximately 0.4987.
c. At least 29:
z = (29 - 33) / 4 = -1
The cumulative probability of z being less than -1 is approximately 0.1587. So the probability that a randomly selected x-value from the distribution is at least 29 is approximately 1 - 0.1587 = 0.8413.
d. At most 21:
z = (21 - 33) / 4 = -3
The cumulative probability of z being less than -3 is very close to 0. So the probability that a randomly selected x-value from the distribution is at most 21 is approximately 0.