Question

A set of data is normally distributed with a mean of 54 and a standard deviation of 2.3. What percentage of scores would lie in the interval from 51.7 to 58.6?

Answer 1

Solution:
Given in the question
A set of data is normally distributed with
Mean() = 54
Standard deviation() = 2.3
We need to calculate the percentage of scores would lie in the interval from 51.7 to 58.6 i.e. P(51.7<X<58.6)=P(X<58.6)-P(X<51.7)
Here we will use the standard normal distribution, first, we will calculate Z-score which can be calculated as
Z-score = (X-)/ = (51.7/54)/2.3 = -1
Z-score = (58.6-54)/2.3 = 2
From Z table we found p-value
P(51.7<X<58.6)=P(X<58.6)-P(X<51.7) = 0.9773 - 0.1587 = 0.8186
So there is 81.86% percentage of scores would lie in the interval from 51.7 to 58.6.