Question

find the systolic blood pressure for men that represents: (e) the 90th percentile (f) the bottom 4% 3) a special task force of a military unit requires that the recruits not be too tall or too short. suppose 12% of the applicants are rejected because they are too tall and 18% because they are too short. if the height of an applicant is normally distributed with a mean of 69.4 inches and a standard deviation of 3.5 inches, determine the heights that define whether an applicant is accepted or rejected. 4) if z = 1.5, $\bar{x}$ = 22, and s = 8.5, what is your “x” value? 5) if z = -0.55, $\bar{x}$ = 13.4, and s = 2, what is your “x” value? 6) if you have 20% of your data lying below a z - score, what is that z - score? 7) if you have 34.5% of your data lying above a z - score, what is that z - score?

Explanation:

Step1: Recall z - score formula

The z - score formula is $z=\frac{x-\bar{x}}{s}$, which can be rewritten to solve for $x$ as $x = zs+\bar{x}$.

Step2: Solve for question 4

Given $z = 1.5$, $\bar{x}=22$, and $s = 8.5$. Substitute these values into the formula $x=zs+\bar{x}$.
$x=1.5\times8.5 + 22=12.75+22 = 34.75$

Step3: Solve for question 5

Given $z=-0.55$, $\bar{x}=13.4$, and $s = 2$. Substitute into the formula $x=zs+\bar{x}$.
$x=-0.55\times2+13.4=-1.1 + 13.4=12.3$

Step4: For question 6

If 20% of the data lies below a z - score, we look up the area 0.20 in the standard normal distribution table. The closest value in the table gives a z - score of approximately $z=-0.84$.

Step5: For question 7

If 34.5% of the data lies above a z - score, then the area to the left of the z - score is $1 - 0.345=0.655$. Looking up this value in the standard normal distribution table, the z - score is approximately $z = 0.40$.

Step6: For question 3

For the lower bound: The area to the left of the lower - bound z - score is 0.18. Looking up 0.18 in the standard normal table, the z - score $z_1\approx - 0.92$. Using the z - score formula $z=\frac{x-\bar{x}}{s}$, and solving for $x$ gives $x_1=z_1s+\bar{x}=-0.92\times3.5 + 69.4=-3.22+69.4 = 66.18$ inches.
For the upper bound: The area to the left of the upper - bound z - score is $1 - 0.12 = 0.88$. Looking up 0.88 in the standard normal table, the z - score $z_2\approx1.175$. Using the z - score formula $x_2=z_2s+\bar{x}=1.175\times3.5+69.4 = 4.1125+69.4=73.5125\approx73.51$ inches.

Answer:

1. $x = 34.75$
2. $x = 12.3$
3. $z=-0.84$
4. $z = 0.40$
5. Lower bound height: 66.18 inches, Upper bound height: 73.51 inches