Answer:
, 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:
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
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:
![\mu = 24.4, \sigma = 1.2](https://tex.z-dn.net/?f=%5Cmu%20%3D%2024.4%2C%20%5Csigma%20%3D%201.2)
In this problem, a value x is significantly high if:
![P(X \geq x) = 0.01](https://tex.z-dn.net/?f=P%28X%20%5Cgeq%20x%29%20%3D%200.01)
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
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
![Z = \frac{26.6 - 24.4}{1.2}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7B26.6%20-%2024.4%7D%7B1.2%7D)
![Z = 1.83](https://tex.z-dn.net/?f=Z%20%3D%201.83)
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.