Answer to Question #314046 in Statistics and Probability for Nickname

Question

The average cost per household of owning a brand new car is Php 5,000. Suppose that we randomly selected 40 households, determine the probability that the sample mean for these 40 households is more than Php 5,350. Assume that the variable is normally distributed and the standard deviation is Php 1,230.

SOLVE FOR THE Z-SCORE

Answer 1

We have a normal distribution, $\mu=5000, \sigma=1230, n=40.$

Let's convert it to the standard normal distribution,

$z=\cfrac{\bar{x}-\mu}{\sigma/\sqrt{n}}$ =

$=\cfrac{5350-5000}{1230/\sqrt{40}}=1. 80,$

$P(\bar{X}>5350)=1-P(\bar{X}<5350)=\\ =1-P(Z<1.80)=$

$=1-0.9641=0.0359$ ​(from z-table)