Get the answers you need at Westonci.ca, where our expert community is dedicated to providing you with accurate information. Experience the ease of finding accurate answers to your questions from a knowledgeable community of professionals. Experience the ease of finding precise answers to your questions from a knowledgeable community of experts.

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]