Isaac is a professional swimmer who trains, in part, by running. She would like to estimate the average number of miles she runs in each week. For a random sample
of 20 weeks, the mean is
x
= 17.5 miles with standard deviation s = 3.8 miles. Find
a 99% confidence interval for the population mean number of weekly miles Isaac runs.
(a) 15.01 to 19.99 miles (b) 15.07 to 19.93 miles
(c) 15.34 to 19.66 miles (d) 15.31 to 19.69 miles
(e) 15.08 to 19.92 miles

(b) 15.07 to 19.93 miles

Step-by-step explanation:

Confidence interval is written in the form,

(Sample mean - margin of error, sample mean + margin of error)

The sample mean, x is the point estimate for the population mean.

Margin of error = z × s/√n

Where

s = sample standard deviation = 3.8

n = number of samples = 20

From the information given, the population standard deviation is unknown and the sample size is small, hence, we would use the t distribution to find the z score

In order to use the t distribution, we would determine the degree of freedom, df for the sample.

df = n - 1 = 20 - 1 = 19

Since confidence level = 99% = 0.99, α = 1 - CL = 1 – 0.99 = 0.01

α/2 = 0.02/2 = 0.005

the area to the right of z0.005 is 0.025 and the area to the left of z0.025 is 1 - 0.005 = 0.995

Looking at the t distribution table,

z = 2.861

Margin of error = 2.861 × 3.8/√20

= 2.43

the lower limit of this confidence interval is

17.5 - 2.43 = 15.07 miles

the upper limit of this confidence interval is

17.5 + 2.43 = 19.93 miles

your answer is a

step-by-step explanation:

4*4*4*4=256

it will cost dollar $540

the living room is 3 which means if we divide 325/3 then we get $108

therefore if 1part of the living room cost $108 and 3 costed $325

then total 5part of the room will cost 5*108 which gives $540