Use the confidence level and sample data to find a confidence interval for estimating the population mean. round your answer to the same number of decimal places as the sample mean.
of the packages received by a parcel service, 37 packages were randomly selected. the sample has a mean weight of 17.0 pounds and a standard deviation of 3.3 lb. what is the 95% confidence interval for the true mean weight, μ, of all packages received by the parcel service?
a)15.7 lb< μ< 18.3 lb
b)15.9 lb< μ< 18.1 lb
c)15.6 lb< μ< 18.4 lb
d)16.1 lb< μ< 17.9 lb

B) 15.9 lb < μ < 18.1 lb

Step-by-step explanation:

Confidence interval = mean ± margin of error

CI = μ ± ME

The mean is μ = 17.0.

Margin of error = critical value × standard error

ME = CV × SE

n > 30, so we can approximate CV with a normal distribution.  At P = 95%, CV = 1.96.

SE = σ / √n

SE = 3.3 / √37

SE = 0.54

So the margin of error is:

ME = 1.96 × 0.54

ME = 1.1

So the confidence interval is:

CI = 17.0 ± 1.1

CI = (15.9, 18.1)

the amount you'd have after the work is $1.55

step-by-step explanation:

you subtract 95% from $31.

part a

in a city ice cream is mostly only bought when the temperature is very high as ice cream cools down our

so we can derive the conclusion that as the temperature increases the number of ice creams sold also increases

part b

y intercept is 5

slope is 2

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