Question

The U.S. Census Bureau announced that the median sales price in 2014 for new homes was $282,800 and the mean sales price was $345,800. The standard deviation was $90,000.
a. If you selected a random sample of 100 homes sold, what is the probability that the sample mean would be less than $370,000?
b. If you selected a random sample of 100 homes sold, what is the probability that the sample mean would be between $350,000 and $360,000?

Answer 1

Answer:

a) 0.9964

b) 0.3040

Step-by-step explanation:

Given data:

standard deviation =  $90,000

Mean sales price =$345,800

sample mean =  $370,000

Total number of sample = 100

calculate z score for [/tex](\bar x = 370000)[/tex]

$z = \frac{\bar x - \mu}{\frac{\sigma}{\sqrt{n}}}$

$z = \frac{370000 - 345800}{\frac{90000}{\sqrt{100}}}$

z = 2.689

P(x<370000) = P(Z<2.689)

FROM STANDARD NORMAL DISTRIBUTION TABLE FOR Z P(Z<2.689) = 0.9964

B)

calculate z score for (\bar x = 350000)

$z = \frac{\bar x - \mu}{\frac{\sigma}{\sqrt{n}}}$

$z = \frac{350000 - 345800}{\frac{90000}{\sqrt{100}}}$

z = 2.133

$P(350000 \leq \bar x \leq 365000) = P(0.467 \leq Z\leq 2.133) = P(Z\leq 2.133) - P(Z\leq 0.467)$

FROM NORMAL DISTRIBUTION TABLE Z VALUE FOR

$P(Z\leq 2.133) = 0.9836$

$P(Z\leq 0.467) = 0.6796$

SO,  = 0.9836 - 0.6796 = 0.3040