Question

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

Answer 1

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.

Answer 2

C.) 217

Explanation:

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

We look for a z-score below which
`96\%` of the population lie.

Reading from the z-table as shown in the attachment, this z-value corresponds to
`1.74`.

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

`Z=(x-\mu)/(\sigma)`

where
`\mu=200,\sigma=10,and\:z=1.74`

We substitute and solve for x.

`1.74=(x-200)/(10)`

`x-200=17.4`

`x=200+17.4`

`x=217.4`

`x\approx217`

The correct choice is C