Suppose a ward of Wichita contains 4,500 detached houses. A sample of 100 houses

Question

Suppose a ward of Wichita contains 4,500 detached houses. A sample of 100 houses is selected randomly and evaluated by an appraiser. Suppose that the mean appraised value of a house in this ward for all houses is actually $925,000, with a standard deviation of $85,500.

(a) What is the probability that the sample average is greater than $950,000?

(b) What is the probability that the sample average is greater than $940,000?

(c) What is the probability that the sample average is greater than $930,000?

(d) What is the probability that the sample average is greater than $920,000?

Explanation / Answer

Mean ( u ) =925000
Standard Deviation ( sd )= 85500/ Sqrt(n) = 8550
Number ( n ) = 100
Normal Distribution = Z= X- u / (sd/Sqrt(n) ~ N(0,1)                  
a.
P(X > 950000) = (950000-925000)/85500/ Sqrt ( 100 )
= 25000/8550= 2.924
= P ( Z >2.924) From Standard Normal Table
= 0.0017                  
b.
P(X > 940000) = (940000-925000)/85500/ Sqrt ( 100 )
= 15000/8550= 1.7544
= P ( Z >1.7544) From Standard Normal Table
= 0.0397                  
c.
P(X > 930000) = (930000-925000)/85500/ Sqrt ( 100 )
= 5000/8550= 0.5848
= P ( Z >0.5848) From Standard Normal Table
= 0.2793                  
d.
P(X > 920000) = (920000-925000)/85500/ Sqrt ( 100 )
= -5000/8550= -0.5848
= P ( Z >-0.5848) From Standard Normal Table
= 0.7207                  

Related Questions

Navigate

• Browse (All)
• Browse S
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.