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The scores of individual students on the American College Testing (ACT) composite college entrance examination have a normal distribution with mean 18.6 and standard deviation 5.9. What is the probability that the mean score of an SRS of 40 students chosen from all those taking the test is 21 or higher? Round your percentage to 2 decimals.

Sagot :

Given data

*The given mean is

[tex]\mu=18.6[/tex]

*The given standard deviation is

[tex]\sigma=5.9[/tex]

The value of the z score is calculated as

[tex]z=\frac{x-\mu}{\sigma}[/tex]

Substitute the values in the above expression as

[tex]\begin{gathered} z=\frac{21-18.6}{5.9} \\ =0.41 \end{gathered}[/tex]

The probability that the mean score of an SRS of 40 students chosen from all those taking the test is 21 or higher is given as

[tex]\begin{gathered} P(Z\ge21)=P(X\ge0.41) \\ =1-P(X<0.41) \end{gathered}[/tex]

The corresponding probability is evaluated by the table.

Substitute the values in the above expression as

[tex]\begin{gathered} P(Z\ge21)=1-0.6591 \\ =0.34 \end{gathered}[/tex]

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