. The U.S. Census Bureau announced that the median sales price of new houses sol

Question

. The U.S. Census Bureau announced that the median sales price of new houses sold in 2014 was $282,000, and the mean sales price was $345,800. Assume that the standard deviation of the prices is $90,000.                                                                                                                                        

a. If you select samples of n=4, describe the shape of the sampling distribution of

b. If you select samples of n=100, describe the shape of the sampling distribution of

c. If you select a random sample of n=100, what is the probability that the sample mean will be less than $370,000?

d. If you select a random sample of n=100, what is the probability that the sample mean will be between $ 350,000 and $ 365,000?

Explanation / Answer

Let X be the random variable that sales price of new houses sold in 2014.

Given that X has mean $345,800 and sd = $90,000 and meadian = $282,000

a. If you select samples of n=4, describe the shape of the sampling distribution of

Here we see that,

Median < mean

Since the samples with sample sizes of n=4 is fairly extremely small (much less than 30), in this case the shape of the sampling distribution of x would be looked skewed to the right and is very spread out.

b. If you select samples of n=100, describe the shape of the sampling distribution of

The sample is an amount greater than 30. This means, based on the CentralLimit Theorem the distribution is approximately normal or centred over the same mean of $345,800.

c. If you select a random sample of n=100, what is the probability that the sample mean will be less than $370,000?

n = 100

Here we have to find P(Xbar < 370000).

We see that the distribution of Xbar is normal with mean = 345,800 and standard deviation = 90,000/sqrt(100) = 9000.

Convert xbar = 345,800 into z-score.

z-score s defined as,

z = (Xbar - mean) / sd

z = (3,45,800 - 370,000) / 9000 = -2.69

Now we have to find P(Z < -2.69)

This probability we can find in EXCEL.

syntax :

=NORMSDIST(z)

where z is z-score.

P(Z < -2.69) = 0.0036

The probability that the sample mean will be less than $370,000 is 0.0036.

d. If you select a random sample of n=100, what is the probability that the sample mean will be between $ 350,000 and $ 365,000?

Now we have to find P(3,50,000 < Xbar < 3,65,000)

Now convert xbar = 350,000 and xbar = 365,000 into z-score

z = (350,000 - 345,800) / 9000 = 0.47

z = (365,000 - 345,800) / 9000 = 2.13

Now we have to find (0.47 < Z < 2.13)

P(0.47 < Z < 2.13) = P(Z < 2.13) - P(Z < 0.47)

= 0.9836 - 0.6796

= 0.3039

The probability that the sample mean will be between $ 350,000 and $ 365,000 is 0.3039.

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