The grades on the last history exam had a mean of 80%. assume the population of grades on history exams is known to be distributed normally, with a standard deviation of 5%. approximately what percent of students earn a score between 70% and 80%?

To calculate for the percent of students who earned scores of 70% and 80%, we compute for the z-scores first and translate the answer to percentiles.

z-scores calculation:

                    (70%)       z-score = (70 – 80)/5 = -2

This translates to percentile of 2.28%

                    (80%)       z-score = (80 – 80)/5 = 0

This translates to percentile of 50%.

Subtracting the percentiles will give us the final answer of 47.72%. 

the  office resources  are the human, physical and financial  resources  which are utilized for performing all the activities in order to achieve the organizational objectives.

happy to

pls mark as brainliest

d. 24+-0.114

step-by-step explanation:

the question requires us to calculate the approximate 95% confidence interval for the mean number of ounces of ketchup per bottle in the sample given the following statistics;

sample mean = 24

standard deviation, σ = 0.4

sample size , n = 49

the confidence interval for a population mean is calculated using the formula;

sample mean ± z-score*(σ/sqrt(n))

the z-score associated with a 95% confidence interval, from the standard normal tables, is;

1.96

substituting these values into the formula;

24 ± 1.96*(0.4/7)

= 24 ± 0.112

from the alternatives given, choice d is closest to the above expression;