Scores on a certain test are normally distributed with a variance of 88. a researcher wishes to estimate the mean score achieved by all adults on the test. find the sample size needed to assure with 95 percent confidence that the sample mean will not differ from the population mean by more than 3 units.

To solve for the problem, the formula is:

n = [z*s/E]^2

where:

z is the z value for the confidence interval

s is the standard deviation

E is the number of units

So plugging that in our equation, will give us:

= [1.96*9.38/3]^2

= (18.3848/3)^2

= 6.1284^2

= 37.5 or 38

ruth is 6 years old

step-by-step explanation:

6 + 24 = 30,   so 6/30 = .2  

if you do 1/5 it also equals .2, so that is the answer.