Question

Suppose that the sitting​ back-to-knee length for a group of adults has a normal distribution with a mean of mu equals 23.1μ=23.1 in. and a standard deviation of sigma equals 1.2σ=1.2 in. These data are often used in the design of different​ seats, including aircraft​ seats, train​ seats, theater​ seats, and classroom seats. Instead of using 0.05 for identifying significant​ values, use the criteria that a value x is significantly high if​ P(x or ​greater)less than or equals≤0.01 and a value is significantly low if​ P(x or ​less)less than or equals≤0.01. Find the​ back-to-knee lengths separating significant values from those that are not significant. Using these​ criteria, is a​ back-to-knee length of 25.225.2 in. significantly​ high?

Answer 1

Answer:

The the​ back-to-knee lengths separating significant values from those that are not significant are X=20.3 and X=25.9.

A value X=25.2 is within the expected values (under the high critical value), so it is not significantly high.

Step-by-step explanation:

We have the distance back-to-knee described by a normal distribution with mean 23.1 in. and standard deviation of 1.2 in.

We have to calculate the value for x, so that it has only a expected probability of 0.01 of having a distance bigger than x.

That is:

$P(X>x)<0.01$

We calculate this using the standard normal distribution.

We get that the z-value so that P(z>z_c)<0.01 is z=2.33.

We can use this result for the back-to-knee distance distribution.

$X=\mu+z\sigma=23.1+2.33*1.2=23.1+2.8=25.9$

We can calculate the same for the critical value in the left tail.

Then, for P(z<z_c)<0.01, the value of z is z=-2.33.

Then we have:

$X=\mu+z*\sigma=23.1-2.33*1.2=23.1-2.8=20.3$

A value X=25.2 is within the expected values (under the high critical value), so it is not significantly high.