Question

The U.S. Census Bureau announced that the mean sale price of new houses sold in 2011 was $267,900. Assume that the standard deviation of the prices is $90,000.

(a) If you select a random sample of n = 100, what is the probability that the sample mean will be less than $300,000?

(b) If you select a random sample of n = 100, what is the probability that the sample mean will be between $275,000 and $290,000?

(c) Between what two values symmetrically distributed around the mean are 80% of the sample mean home prices, when n = 100?

Answer 1

Solution :

Given that ,

mean = = 267900

standard deviation = = 90000

n = 100

= = 267900  

= / n = 90000/ 100 = 9000

a) P( <300000 ) = P(( - ) / < (300000 - 267900) /9000 )

= P(z < 3.57)

= 0.9998

probability = 0.9998

b)

P(275000< < 290000 )  

= P[(275000 - 267900) /9000 < ( - ) / < ( 290000 - 267900) /9000 )]

= P( 0.79< Z < 2.46 )

= P(Z < 2.46 ) - P(Z < 0.79 )

= 0.9931 - 0.7852 = 0.2078

probability = 0.2078

c) 80%

P(-z Z z) = 0.80

P(Z z) - P(Z -z) = 0.80

2P(Z z) - 1 = 0.80

2P(Z z) = 1 + 0.80 = 1.80

P(Z z) = 1.80 / 2 = 0.90

P(Z 1.28) = 0.90

z = -1.28 , +1.28

z = -1.28

= z * + = -1.28 * 9000+267900 = 256380

z = 1.28

= z * + = 1.28 * 9000+267900 = 279420

Answer = Between two values is 256380 and 279420.