Answer to Question #136174 in Statistics and Probability for Princess

Question

A restaurant offers a 1kg steak as part of a promotion. Assume weights of the steaks are approximately normally distributed with a standard deviation of 25 grams.
X= weights of the steaks in grams
X bar= average weights of the steaks in grams
Assume that the average weights on 5 restaurants has been investigated.

Calculate the 75th percentile of the average weight of steaks for the 5 restaurants.

Answer 1

solution

let $X_{0.75}$ be the weight of the steak at the 75th percentile. $75\%$ of the steaks will weigh less than $X_{0.75}$ . Mean weight of steak served $\bar X =1\ kg =1000\ grams$ and standard deviation $\sigma =25$

The corresponding $Z- value$ is

$0.67$

$$Z=\frac{X_{0.75}-\bar X}{\sigma}$$

$$0.67=\frac{X_{0.75}-1000}{25}$$

$$X_{0.75}=0.67*25+1000 = 1016.75$$

answer: the 75th percentile of the average weights is 1016.75 grams or 1.01675 kg