Question

the systolic blood pressures of women aged 18 to 24 are normally distributed with a mean of 119.4 and a standard deviation of 14.9. what is the systolic blood pressure interval for the central 90% of all women between the ages of 18 - 24?
a. (85.1464, 118.3569)
b. (110.2456, 123.9877)
c. (94.8917, 143.9083)
d. (97.9510, 140.8486)

Explanation:

Step1: Find the z - scores

For the central 90% of a normal distribution, the area in the two - tails is $1 - 0.90=0.10$. So the area in each tail is $\frac{0.10}{2}=0.05$. The z - score corresponding to an area of $0.05$ in the left - tail is $z_1=- 1.645$ and the z - score corresponding to an area of $0.95$ (since $0.5 + 0.45$) in the right - tail is $z_2 = 1.645$.

Step2: Use the z - score formula

The z - score formula is $z=\frac{x-\mu}{\sigma}$, where $x$ is the value from the data set, $\mu$ is the mean, and $\sigma$ is the standard deviation. We want to solve for $x$, so $x=\mu+z\sigma$. Given $\mu = 119.4$ and $\sigma=14.9$.
For the lower bound: $x_1=\mu+z_1\sigma=119.4+( - 1.645)\times14.9=119.4-24.5105 = 94.8895\approx94.8917$.
For the upper bound: $x_2=\mu+z_2\sigma=119.4 + 1.645\times14.9=119.4+24.5105=143.9105\approx143.9083$.

Answer:

C. (94.8917, 143.9083)