Question

question 4 of 10
suppose a normal distribution has a mean of 98 and a standard deviation of 6. what is ( p(x geq 86) )?

a. 0.16
b. 0.475
c. 0.84
d. 0.975

Explanation:

Step1: Calculate the z-score

The formula for the z - score is $z=\frac{x - \mu}{\sigma}$, where $\mu = 98$, $\sigma=6$ and $x = 86$.
Substitute the values into the formula: $z=\frac{86 - 98}{6}=\frac{- 12}{6}=- 2$.

Step2: Use the properties of the normal distribution

In a normal distribution, the total area under the curve is 1. We know that the area to the left of $z=- 2$ can be found using the standard normal table or the empirical rule. The empirical rule states that about 95% of the data lies within 2 standard deviations of the mean ($\mu\pm2\sigma$), so the area outside of this range (less than $\mu - 2\sigma$ or greater than $\mu+2\sigma$) is $1 - 0.95 = 0.05$. Since the normal distribution is symmetric, the area to the left of $z=-2$ is $\frac{0.05}{2}=0.025$.
We want to find $P(x\geq86)$, which is equal to $1 - P(x\lt86)$. And $P(x\lt86)$ is the area to the left of $x = 86$ (or $z=-2$), which is 0.025. So $P(x\geq86)=1 - 0.025 = 0.975$.

Answer:

D. 0.975