Question

Can you please help with 2,3,4 I am so confused!! Please show work as well

10 out of 10 points Question 1 A population has a mean of 60 and a standard deviation of 30. Samples of size 16 are randomly selected. Calculate the standard deviation of the sample distribution of X. Round off to 3 decimal places. 15 out of 15 points Question 2 Samples of size 16 are drawn from a population. The sampling distribution for X has a standard deviation of 0.25. Find the standard deviation of the population. Round off to three decimal places. O out of 15 points Question 3 A population has a probability distribution as follows. X PO 1 02 20.5 3 0.3 A sample of 2 is drawn and its mean, X, calculated. Find the probability - 2.0 15 out of 15 points Question 4 Tires are found to have a mean life of 40.000 miles. The standard deviation is 8000. A sample of 400 is chosen. Find the probability that 40,500 < x 15 out of 15 points Question 5 Tires are found to have a mean life of 800 hours. The standard deviation is 50. A sample of 100 is chosen. Find the probability that 40 800 < X <812. 15 out of 15 points Question 6 A sample of size 64 is chosen from a population. The sample mean is 50 and the standard deviation is 5. Find the upper limit of the 85% confidence interval for the population mean. Round off to three decimal places. Question 7 O out of 15 points A sample of size 64 is chosen from a population. Assume the probability distribution is normal. If the mean of the sample is 110 and the standard deviation is 6, find the lower bound of the 99% confidence interval. Round off to three x decimal places.

Answer 1

Solution :- Ø Sample Size =N- 16 Sampling distribution oft has Standard deviation of 10.26 of Standard deviation of population to thos (116)(0-25 =1 1-000 Road to three decimal Reason :- If x ~N (402) then Fun (4 (5)?)> Sampling distibution glas CSScanned with CamScannel

Solution Population probability distribution X P(x) 0.2 02 +0.5 to 3 - = 1 > Salisfied Condition 2 O's Unity. 3 03 Ex.for Sample of 2 is drown and its mean is a Mean F(x) To find probability of TX = 2 P(X=2.0) = 0:3 Solution l= yo,000 8000 let x n N (40,000, 800 n= yoo ANNO 40,000 rovo yoo :)? / - 400 then P (40,500 <*) P(z> U SOO-40,000 yoo X-H = Z = = P(Z > 125) 1-P1231.25) FI - 0.8944 = 0.1056 p(40500 <x)= 0.105€ CScanned with CamScanner