Suppose the test scores on an exam show a normal distribution with a mean of 82 and a standard deviation of 4.
a. Within what range do about 95% of the scores fall?
b. About what percent of the scores are between 78 and 90?

a) Between 74 and 90

b) 81.86% of the scores are between 78 and 90

Step-by-step explanation:

a) According to the Empirical Rule, in a normal distribution, 95% of the data are within 2 standard deviations of the mean. Therefore, the range that 95% of the scores fall in is between 82-4(2)=74 and 82+4(2)=90.

b) The percent of scores that are between 78 and 90 is: normalcdf(lower,upper,μ,σ) = normalcdf(78,90,82,4) = 0.8185946784 ≈ 0.8186, or about 81.86% of the scores.

the answer is d. (1.2, -4.7) on edgn

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