For a certain urban area, in a random sample of 7 months, an average of 28 mail carriers were bitten by dogs each month. The standard deviation of the sample was 3.5. Find the 99% confidence interval of the true mean number of mail carriers who are bitten by dogs each month. Assume the variable is normally distribute. a. (29.38)
b. (25, 33)
c. (23,33)
d. (23.35)

C

Step-by-step explanation:

Mean = 28

S = 3.5

N = 7

We use the t distribution here.

Level of significance = 99%

Degree of freedom = 7-1 = 6

Using the t table at 99%, and df of 6

T' = t tabulated = 3.7

Confidence interval =[ (mean - t's)/√n , mean + t's)/√n]

= 28-3.7*3.5/√7 , 28+3.7*3.5/√7

= 23.11, 32.89

When approximated

= 23, 33 after rounding to the nearest integer

Therefore option c is the answer

my acc plz

step-by-step explanation:

the correct answer is x/3+10≥32

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