Question

The scores on an exam are normally distributed, with a mean of 74 and a standard deviation of 7. What percent of the scores are less than 81?
A. 84%

B. 13.5%

C. 50%

D. 16%

Answer 1

A normal distribution with a mean of 74 and a standard deviation of 7.
Mean + 1 SD = 74 + 7 = 81
Less than 81 :  50 % + 34 % = 84 %
Answer:
A ) 84 % 

Answer 2

Answer:

84% of the scores are less than 81.

Step-by-step explanation:

Here,

The given distribution is in normal distribution.

Mean = 74

Standard deviation = 7

Approximately 68% of the data falls within one standard deviation of the mean (or between the mean -one times the standard deviation, and the mean + 1 times the standard deviation).

As 81 = 74+7 = μ+1σ

So, $\dfrac{64}{2}=34\%$ of scores will be between 74 and 81.

As 74 is the mean, so 50% of score will lie below 74.

So total of 50+34=84% of the score will be less than 81.