No steps, only answer required ( in 4 decimal places) 1) The mean monthly rent f

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No steps, only answer required ( in 4 decimal places)
1) The mean monthly rent for a non-bedroom apartment without a doorman in Manhattan is $2648. Assume the standard deviation is $500. A real estate firm samples 100 apartments. What is the probability that the average rent of the sample is more than $2704?
2) A survey bills in a certain city have mean $89.20. Assume the bills are normally distributed with standard deviation $11.71. A sample of 48 bills was selected for an adult. Find the 46 percentile for the sample mean
3) A survey among freshman at a certain university revealed that the number of hours spent studying the week before final exams was normally distributed with mean 25 and standard deviation 15. A sample of 36 students was selected. What is the probability that the average time spent studying for the sample was between 26.5 and 30 hours studying?
4) Electricity bills in a certain city have mean $89.20. Assume the bills are normally distributed with standard deviation $11.71. A sample of 48bills was selected for an adult. Find the 46 percentile for the sample mean. No steps, only answer required ( in 4 decimal places)
1) The mean monthly rent for a non-bedroom apartment without a doorman in Manhattan is $2648. Assume the standard deviation is $500. A real estate firm samples 100 apartments. What is the probability that the average rent of the sample is more than $2704?
2) A survey bills in a certain city have mean $89.20. Assume the bills are normally distributed with standard deviation $11.71. A sample of 48 bills was selected for an adult. Find the 46 percentile for the sample mean
3) A survey among freshman at a certain university revealed that the number of hours spent studying the week before final exams was normally distributed with mean 25 and standard deviation 15. A sample of 36 students was selected. What is the probability that the average time spent studying for the sample was between 26.5 and 30 hours studying?
4) Electricity bills in a certain city have mean $89.20. Assume the bills are normally distributed with standard deviation $11.71. A sample of 48bills was selected for an adult. Find the 46 percentile for the sample mean. No steps, only answer required ( in 4 decimal places)

2) A survey bills in a certain city have mean $89.20. Assume the bills are normally distributed with standard deviation $11.71. A sample of 48 bills was selected for an adult. Find the 46 percentile for the sample mean
3) A survey among freshman at a certain university revealed that the number of hours spent studying the week before final exams was normally distributed with mean 25 and standard deviation 15. A sample of 36 students was selected. What is the probability that the average time spent studying for the sample was between 26.5 and 30 hours studying?
4) Electricity bills in a certain city have mean $89.20. Assume the bills are normally distributed with standard deviation $11.71. A sample of 48bills was selected for an adult. Find the 46 percentile for the sample mean.

Explanation / Answer

1) P(X > 2704) = P((X - mean)/(sd/sqrt(n)) > ((2704 - mean)/(sd/sqrt(n)))

                         = P(Z > (2704 - 2648)/(500/sqrt(100)))

                         = P(Z > 1.12)

                         = 1 - P(Z < 1.12)

                         = 1 - 0.8686

                         = 0.1314

b) P(X < x) = 0.46

or, P((X - mean)/(sd/sqrt(n)) < (x - 89.2)/(11.71/sqrt(48))) = 0.46

or, P(Z < ((x - 89.2)/(11.71/sqrt(48))) = 0.46

or, (x - 89.2)/(11.71/sqrt(48)) = -0.1

or, x = -0.1 * (11.71/sqrt(48)) + 89.2

or, x = 89.03

c) P(26.5 < X < 30)

= P((26.5 - 25)/(15/sqrt(36)) < Z < (30 - 25)/(15/sqrt(36)))

= P(0.6 < Z < 2)

= P(Z < 2) - P(Z < 0.6)

= 0.9772 - 0.7257

= 0.2515

d) P(X < x) = 0.46

or, P((X - mean)/(sd/sqrt(n)) < (x - 89.2)/(11.71/sqrt(48))) = 0.46

or, P(Z < ((x - 89.2)/(11.71/sqrt(48))) = 0.46

or, (x - 89.2)/(11.71/sqrt(48)) = -0.1

or, x = -0.1 * (11.71/sqrt(48)) + 89.2

or, x = 89.03

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