Question

Suppose a normal distribution has a mean of 98 and a standard deviation of

6. What is P(x ≤ 104)?
O A. 0.975
O B. 0.025
O c. 0.16
D. 0.84

Answer 1

0.84

Explanation:

The probability of a value being greater than 104 in a normal distribution with a mean of 98 and a standard deviation of 6 can be calculated using the z-score formula. The z-score for 104 is (104-98)/6 = 1. The probability of a value being greater than 104 is equal to 1 minus the cumulative probability of the Z-score, which is 0.84.