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?

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:

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.

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:

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

answer: b.

step-by-step explanation:

the y coordinate for point b is 0, and answer b is the only one with that y coordinate.

answer:   35 % is the probability of a late departure or an early arrival.

step-by-step explanation:

let l represents the late departure and e represents the early departure.

then, according to the question,

p(l) = 12 %,   p(e) = 27 % and = 4 %

and, we have to find out,

but, we know that,

which is the required equation that will use to find the probability of a late departure or an early departure.

therefore,   the probability of a late departure or an early arrival =   35 %