Question

A test was given to 120 students, and the scores approximated a normal distribution If
the mean score was 72 with a standard deviation of 7, approximately what percent of the
scores were 65 or higher?

Answer 1

Answer:

84.13% aka. 101/120 students

Step-by-step explanation:

P(X>65) = normalcdf(65,100,72,7) = 0.8413130544 ≈ 0.8413 ≈ 84.13%

Therefore, about 84.13% of the scores were 65 or higher, meaning that about 101 students got a score of 65 or higher.

Answer 2

84% of students have a score of 65 or higher.

What is Normal Distribution?

'Normal distribution is a continuous probability distribution wherein values lie in a symmetrical fashion mostly situated around the mean.'

According to the given problem,

For this sum, the mean and standard deviation are mentioned.

Mean = 72

S.D = 7

We will be using the bell curve of normal distribution for solving this sum.

Area under the curve that represents more than 65:

=  34% + 34% + 13.50% + 2.35% + 0.15%

=  84%

Hence, we can conclude that 84% of the scores were 65 or higher.

Learn more about normal distribution: https://brainly.com/question/13759327

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