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suppose a normal distribution has a mean of 62 and a standard deviation of4. What is the probability that a data value is between 54 and 66? Round youranswer to the nearest tenth of a percent

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Solution

For this case we have the following info given:

[tex]X\approx N(62,4)[/tex]

and we want to find this probability:

P(54And we can use the z score and we have:

[tex]z=\frac{54-62}{4}=-2[/tex][tex]z=\frac{66-62}{4}=1[/tex]

And we have this:

[tex]P(54And the final answer would be 81.9%

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