Question

question #6
a normally distributed population has a mean of 500 and a standard deviation of 150.
determine that value such that 95% of all other values are larger than it.
287
278
235
253

question #7
a population is normally distributed with a mean of 67 and a standard deviation of

1. determine the value that is larger than 95% of all other values.

95.91
95.09
92.17
94.96

Explanation:

Step1: Find the z - score

For a normal distribution, if we want to find the value \(x\) such that \(P(X>x) = 0.95\), then \(P(X\leq x)=1 - 0.95=0.05\). Looking up in the standard normal distribution table (z - table), the z - score \(z\) corresponding to a cumulative probability of \(0.05\) is approximately \(z=-1.645\) (for the left - tailed case). For the case of finding the value \(x\) such that \(P(X > x)=0.05\) (right - tailed), the z - score \(z = 1.645\).

Step2: Use the z - score formula

The z - score formula is \(z=\frac{x-\mu}{\sigma}\), where \(\mu\) is the mean and \(\sigma\) is the standard deviation.

For Question #6:

Given \(\mu = 500\), \(\sigma=150\), and \(z=-1.645\) (since \(P(X > x)=0.95\) implies \(P(X\leq x) = 0.05\) and we use the left - tailed z - score).
Substitute into the formula \(z=\frac{x-\mu}{\sigma}\):
\(-1.645=\frac{x - 500}{150}\)
Multiply both sides by \(150\): \(x-500=-1.645\times150\)
\(x-500=-246.75\)
Add \(500\) to both sides: \(x=500-246.75 = 253.25\approx253\)

For Question #7:

Given \(\mu = 67\), \(\sigma = 17\), and \(z = 1.645\) (since \(P(X>x)=0.05\) and we use the right - tailed z - score)
Substitute into the formula \(z=\frac{x-\mu}{\sigma}\):
\(1.645=\frac{x - 67}{17}\)
Multiply both sides by \(17\): \(x-67=1.645\times17\)
\(x-67 = 27.965\)
Add \(67\) to both sides: \(x=67 + 27.965=94.965\approx94.96\)

Answer:

Question #6: \(253\)
Question #7: \(94.96\)