Question

The winning scores of all college​ men's basketball games in a particular season were approximately normally distributed with mean 79.5 points and standard deviation 13.5 points.

What interval of winning scores would be the central 95% of all winning scores for this​ season?

Answer 1

[ 66 ; 93 ] The interval for 95% of all winning scores of this season

Step-by-step explanation:

We will examine this question with the empirical rule. Empirical rule established, that for a standard normal distribution, of all values of a given population

μ + σ will contains 68.3 %

μ + 2σ will contains 95.5 % and

μ + 3σ will contains 99.7 %

We have to understand that the intervals for that mathematics expressions are:

μ ± 0.5σ or [ μ - 0.5*σ ; μ + 0.5*σ ]

μ ± 1σ or [ μ - σ ; μ + σ ]

μ ± 1.5σ [ μ - 1.5*σ ; μ + 1.5*σ ]

respectively.

Then the interval

[ μ - σ ; μ + σ ] will contains 95% of winning scores.

In this case

μ - σ = 79.5 - 13.5 = 66 and

μ + σ = 79.5 + 13.5 = 93

So our interval is

[ 66 ; 93 ]