A competitive cheer team allows only the top 4% of athletes who try out to be part of the team. If the team tryout scorecard has a mean of 200 and a standard deviation of 10, which of the following can be used as a minimum qualifying score to join the cheer team?

A.) 8
B.) 192
C.) 217
D.) 202

Question

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kudzordzifrancis kudzordzifrancis · Answer 1 · 2020-01-24T18:18:18+01:00

Answer:

C.) 217

Step-by-step explanation:

If the top 4% of athletes are allowed to be part of the team, then the bottom [tex]1-4\%=96\%[/tex] will not qualify.

We look for a z-score below which [tex]96\%[/tex] of the population lie.

Reading from the z-table as shown in the attachment, this z-value corresponds to [tex]1.74[/tex].

We now use the z-score formula to find the required minimum qualifying score.

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

where [tex]\mu=200,\sigma=10,and\:z=1.74[/tex]

We substitute and solve for x.

[tex]1.74=\frac{x-200}{10}[/tex]

[tex]x-200=17.4[/tex]

[tex]x=200+17.4[/tex]

[tex]x=217.4[/tex]

[tex]x\approx217[/tex]

The correct choice is C

hypotenuse314 hypotenuse314 · Answer 2 · 2021-01-15T23:27:05+01:00

Answer:

C.) 217 (I agree)

Step-by-step explanation (just like above):

Think about it, you're looking for the top 4% so the score has to be better than about 96% of everyone else who tried out for the team.

1.74 matches up with 95.91 on the z score table.

Following the equation: z = raw score - mean over over standard deviation

z is 1.74

so the equation should look something like 1.74 = x - 200/ 10

Solve for x and get 217.4 which can be rounded to 217 which is your answer.