Suppose that the sitting back-to-knee length for a group of adults has a normal distribution with a mean of mu equals 23.3 in. and a standard deviation of sigma equals 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 equals0.01 and a value is significantly low if P(x or less)less than or equals0.01. 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 25.6 in. significantly high?

The back-to-knee lengths separating significant values from those that are not significant are 20.2 in. and 26.4 in.

Using these criteria, a back-to-knee length of 25.6 in. is not significantly high, as is within the not-significant values.

Step-by-step explanation:

We have a normal distribution for the sitting back-to-knee length with mean 23.3 in. and standard deviation 1.2 in.

We have to calculate the critical values that limit the 1% right tail and 1% left tail.

This can be done looking for a standard normal distribution table for the values that satisfy:

This values for z are z=±2.576.

Then, we can calculate the bounds for our sitting back-to-knee length distribution as:

the range of a function is the range of y-values for which it is defined, or exists. given that we are only looking at the y-values, we can already discount the first two answers as they deal with the domain of the function (range of x-values).

now, what we need to do is find the smallest and highest y-values for which the function is defined. if you'd like, you could take a ruler and start at the very bottom of the graph, find the lowest point, then move the ruler higher and higher until you reach the highest point. doing so either in your head or with an actual ruler, we can see that the lowest y-value is -5 and the highest y-value is -1.

the question now remains as to whether they are included in the range (fourth answer) or excluded (third answer). the key here is to recognise whether they are marked with an open or a closed circle. since both (-2, -5) and (2, -1) are marked with a closed circle, they are points that exist on the graph and are thus included in the range (note that if they were marked with an open circle like (4, -2) then that would signify that there is a break in the graph at that point).

thus, the fourth answer, (y | -5 ≤ y ≤ -1), is correct.

also note that usually it is good practice to scan your eye inbetween the highest and lowest point for any breaks in the graph - we do have one here at (4, -2), however -2 is still included in the range because there is another point that is included in the graph with the same y-value, ie. point (1, -2).

i hope that but if you require any further explanation don't hesitate to comment below.