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

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.

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Get help: Video

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