What is the probability?

Question

If the selling prices of homes in a neighborhood are normally distributed around a mean of $315,000, with a standard deviation of $25,500, what is the probability of a house in this neighborhood costing between $369,000 and $388,000? Show your work.

Answer 1

Find the z scores of 369,000 and 388,000:

z(369,000) = (369000 - 315000) / 25500 = 1.76

Using the normal probability distribution table, a z score of 1.76 corresponds with a probability of 0.9608

z(388,000) = (388000 - 315000) / 25500 = 2.86

Using the normal probability distribution table, a z score of 2.86 corresponds with a probability of 0.9979

Therefore the probability of a house costing between 369000 and 388000 is (0.9979 - 0.9608) = 0.0371, or 3.71%

Answer 2

Z(369,000) = (369 - 315) / 25.5 = 2.117647
Z(388,000) = (388 - 315) / 25.5 = 2.862745
P($369,000 < price < $388,000) =
P(price < $388,000) - P(price < $369,000) =
P(Z < 2.862745) - P(Z < 2.117647) =
0.997900 - 0.982898 = 0.015002 ≈ 0.015

Edit:
I think the answerer above mis-keyed on
Z(369,000)

Answer 3

For any normal random variable X with mean μ and standard deviation σ, X ~ Normal(μ, σ)

you can translate into standard normal units by:
Z = (X - μ) / σ

where Z ~ Normal(μ = 0, σ = 1). You can then use the standard normal cdf tables to get probabilities.

An applet for finding the values
http://www-stat.stanford.edu/~naras/jsm/FindProbability.html

X ~ Normal( 315000 , 25500 )
Find: P( 369000 < X < 388000 )
= P(( 369000 - 315000 ) / 25500 ) < Z < ( 388000 - 315000 ) / 25500 )
= P( 2.117647 < Z < 2.862745 )
= P( Z < 2.862745 ) - P( Z < 2.117647 )
= 0.9979 - 0.9828975
= 0.01500254