88.9k views
5 votes
Suppose a normal distribution has a mean of 98 and a standard deviation of 6. What is P(x≥92)?

A. 0.16
B. 0.475
C. 0.84
D. 0.975

User Nickel
by
7.3k points

2 Answers

5 votes

Answer:

Explanation:

User Large
by
7.6k points
6 votes

We can solve this problem by standardizing the variable x using the formula:

z = (x - μ) / σ

where μ is the mean and σ is the standard deviation. This formula gives us the number of standard deviations x is from the mean.

Substituting the values given in the problem, we get:

z = (92 - 98) / 6 = -1

This means that 92 is one standard deviation below the mean.

To find P(x ≥ 92), we need to find the area under the normal curve to the right of 92. Since we have standardized the variable, we can use a standard normal distribution table or a calculator to find this area.

Using a standard normal distribution table, we look up the area to the right of -1, which is 0.1587. This means that the area to the left of -1 (which corresponds to x < 92) is 1 - 0.1587 = 0.8413.

Therefore, P(x ≥ 92) = 0.8413, which is closest to option C, 0.84.

User Saruftw
by
8.3k points