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Use technology or a z-score table to answer the question.

Scores on a standardized military exam are normally distributed with a mean of 57 and a standard deviation of 9. Consider a group of 4000 military students.

Approximately how many students will score less than 66 on the test?


A. 200

B. 634

C. 3366

D. 3800

Use technology or a zscore table to answer the question Scores on a standardized military exam are normally distributed with a mean of 57 and a standard deviati class=

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MathPhys

Answer:

C

Step-by-step explanation:

Finding the z-score:

z = (x - μ) / σ

z = (66 - 57) / 9

z = 1

Using a z-score table or calculator:

P(z < 1) = 0.8413

84.13% of 4000 is:

0.8413 (4000) = 3365.2

Rounding up to the nearest whole number, approximately 3366 students will score less than 66.  Answer C.

Using the normal distribution, it is found that approximately 3366 students will score less than 66 on the test.

Normal Probability Distribution

The z-score of a measure X of a normally distributed variable with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex] is given by:

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

  • The z-score measures how many standard deviations the measure is above or below the mean.
  • Looking at the z-score table, the p-value associated with this z-score is found, which is the percentile of X.

In this problem, the mean and the standard deviation are given as follows:

[tex]\mu = 57, \sigma = 9[/tex]

The proportion of students that scored less than 66 is the p-value of Z when X = 66, hence:

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

[tex]Z = \frac{66 - 57}{9}[/tex]

Z = 1

Z = 1 has a p-value of 0.8413.

Out of 4000 students:

0.8413 x 4000 = 3366, which means that option C is correct.

More can be learned about the normal distribution at https://brainly.com/question/24663213

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