Question

Suppose that prices of a certain model of new homes are normally distributed with a mean of $150,000. Find the percentage of buyers who paid:

between $150,000 and $152,400 if the standard deviation is $1200.

Answer 1

The percentage is
`P(x_1 < X < x_2) = 47.7 \%`

Explanation:

From the question we are told that

The population mean is
`\mu = \$ 150000`

The standard deviation is
`\sigma = \$ 1200`

The prices we are considering is
`x_1 = \$150000 \to \ x_2 = \$ 152400`

Given that the price is normally distributed , the percentage the percentage of buyers who paid between $150,000 and $152,400 is mathematically represented as

`P(x_1 < X < x_2) = P((x_1 - \mu)/(\sigma ) < (X - \mu)/(\sigma ) < (x_2 - \mu)/(\sigma ))`

So
`(X - \mu)/(\sigma )` is equal to z (the standardized value of X )

So

`P(x_1 < X < x_2) = P((x_1 - \mu)/(\sigma ) <Z < (x_2 - \mu)/(\sigma ))`

substituting values

`P(x_1 < X < x_2) = P((150000 - 150000)/(1200 ) <Z < (152400 - 150000)/(1200 ))`

`P(x_1 < X < x_2) = P(0<Z < 2)`

`P(x_1 < X < x_2) = P( Z < 2) - P( Z < 0 )`

From the standardized normal distribution table
`P(Z < 2 ) = 0.97725` and

`P(Z < 0) = 0.5`

So

`P(x_1 < X < x_2) = 0.97725 - 0.5`

`P(x_1 < X < x_2) = 0.47725`

The percentage is
`P(x_1 < X < x_2) = 47.7 \%`