Question

Different dealers may sell the same card for different prices. The sale prices for a particular cars are normally distributed with a mean and standard deviation of $26,000 in $2000, respectively. Suppose we select one of these cars at random. X represent the sale price for the selected car. Find P(X<30)

Answer 1

P(X < 30,000) = 0.97725

Explanation:

We are given that the sale prices for a particular cars are normally distributed with a mean and standard deviation of $26,000 in $2000, respectively.

Let X = sale price for the selected car

So, X ~ N(
`\mu = 26,000 , \sigma^(2) = 2000^(2)`)

The z score probability distribution is given by;

Z =
`(X-\mu)/(\sigma)` ~ N(0,1)

where,
`\mu` = population mean

`\sigma` = population standard deviation

So, Probability that selected car has a sale price of less than $30,000 is given by = P(X < 30,000)

P(X < 30,000) = P(
`(X-\mu)/(\sigma)` <
`(30000-26000)/(2000)` ) = P(Z < 2) = 0.97725

Therefore, Probability that selected car has a sale price of less than $30,000 is 0.97725.

Answer 2

0.9772 is the required probability.

Explanation:

We are given the following information in the question:

Mean, μ = $26,000

Standard Deviation, σ = $2000

We are given that the distribution of sale price is a bell shaped distribution that is a normal distribution.

Formula:

`z_(score) = \displaystyle(x-\mu)/(\sigma)`

P(x < 30,000)

`P( x < 30000) = P( z < \displaystyle(30000 - 26000)/(2000)) = P(z<2)`

Calculation the value from standard normal z table, we have,

`P(x < 30000) =0.9772 = 97.72\%`

Thus, 0.9772 is the required probability.