Question

the average heights of men age 18 to 55 in a particular city is 180 cm with a standard deviation of 7.5 cm. if the distribution is normal, then the middle 90% is between what two heights? 164.2 and 193.7 165.3 and 194.7 167.7 and 192.3 160.7 and 199.3

Explanation:

Step1: Find the z - scores

For a normal distribution, the middle \(90\%\) leaves \(10\%\) in the two tails. So, \(5\%\) in each tail.
Looking up in the standard normal table, the z - score corresponding to an area of \(0.05\) in the left tail is \(z=- 1.645\) and the z - score corresponding to an area of \(0.95\) (since \(1 - 0.05=0.95\)) in the right tail is \(z = 1.645\)

Step2: Use the z - score formula \(x=\mu+z\sigma\)

Given \(\mu = 180\) (mean) and \(\sigma=7.5\) (standard deviation)
For \(z=-1.645\):
\(x_1=\mu+z\sigma=180+(-1.645)\times7.5\)
\(x_1=180 - 12.3375=167.6625\approx167.7\)
For \(z = 1.645\):
\(x_2=\mu+z\sigma=180+1.645\times7.5\)
\(x_2=180 + 12.3375=192.3375\approx192.3\)

Answer:

167.7 and 192.3