Suppose that the sitting back-to-knee length for a group of adults has a normal distribution with a mean of mu equals 24.4 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 26.6 in. significantly high?

, which is greater than 0.01. So a back-to-knee length of 26.6 in. is not significantly high.

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean and standard deviation , the zscore of a measure X is given by:

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

In this problem, a value x is significantly high if:

Using these criteria, is a back-to-knee length of 26.6 in. significantly high?

We have to find the probability of the length being 26.6 in or more, which is 1 subtracted by the pvalue of Z when X = 26.6. So

has a pvalue of 0.9664.

1 - 0.9664 = 0.0336

, which is greater than 0.01. So a back-to-knee length of 26.6 in. is not significantly high.

47.28 or 47.2857143

step-by-step explanation:

you just divide 331 by 7 and you get that answer. ( i think so tho)