A supervisor records the repair cost for 22 randomly selected VCRs. A sample mean of $75.50 and standard deviation of $18.07 are subsequently computed. Determine the 99% confidence interval for the mean repair cost for the VCRs. Assume the population is approximately normal.

 The 99% confidence interval for the mean repair cost for the VCRs

(65.801, 85.199)

Step-by-step explanation:

Explanation:-

Mean of the sample(x⁻) = $75.50

Given the standard deviation of the sample (S) =  $18.07

Given the size of the sample 'n' = 22

Degrees of freedom = n-1 =22-1 =21

critical value t₍₀.₀₁, ₂₁₎ = 2.5176

The 99% confidence interval for the mean repair cost for the VCRs

( 75.50 - 9.699 , 75.50 + 9.699 )

(65.801, 85.199)

Final -

The 99% confidence interval for the mean repair cost for the VCRs

(65.801, 85.199)

step-by-step explanation:

12: 00 p.m. to 7: 00 p.m. is a total of 7 hours.

4: 00 p.m. to 7: 00 p.m. is a total of 3 hours.

this means that from 4: 00 p.m to 7: 00 p.m. this represents 3/7 of what was lost from 12: 00 p.m. to 7: 00 p.m.

after knowing this, after doing 11,600 - 6,800 = 4,800

4,800 represents 3/7 of what was lost for 7 hours.

we then divide 4,800 by 3 which gets 1,600.

because we know that 1,600 represents 1/7 of how much water was lost, we just have to multiply by 7, which gets 11,200.   after adding 11,200 to how much water was left at 7: 00 p.m. which was 6,800. 11,200 + 6,800 = 18,000

this means that the pool originally had 18,000 gallons in the pool.

you can also check your work by doing 18,000 - 4(1600), which becomes           1800 - 6400 which gets you 11,600.

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step-by-step explanation: