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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?

Respuesta :

Answer:

[ 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 ]

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