Question

Suppose a normal distribution has a mean of 98 and a standard deviation of6. What is P(x < 110)?A. 0.84B. 0.16C. 0.025O D. 0.975

Answer 1

We know that

• The mean is 98.

,

• The standard deviation is 6.

,

• The given x-value is 110.

First, we find the z-value using the following formula

`Z=(x-\mu)/(\sigma)_{}`

Replacing the given information, we have

`Z=(110-98)/(6)=(12)/(6)=2_{}`

The z-value or z-score is 2.

Then, we use a z-table to find the probability when P(x<110), or P(z<2).

We obtain a probability of 0.97, which approximates to D.

Hence, the probability would be D.