a sample of final exam scores is normally distributed with a mean equal to 25 and a variance equal to 16.

what percentage of scores are between 21 and 29?

Respuesta :

Answer:

The answer is "0.6827".

Step-by-step explanation:

[tex]\bold{X \to N(\mu = 25, \sigma^2 = 16)}[/tex]

[tex]\to P(21 < X < 29) = P(\frac {21-25}{4} < \frac{X- \mu}{\sigma}< \frac{29-25}{4})[/tex]  

                              [tex]= P(-1 < Z< 1)\\\\= P(Z> -1)-P(Z>1)\\\\= 1-P(Z< -1)-P(Z>1)\\\\= 1-P(Z> 1)-P(Z>1)\\\\= 1- 2 \times P(Z>1)\\\\=1-2 \times 0.15866 \\\\ = 1- 0.31732\\\\=0.6827[/tex]

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