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

Answer:

0.9772 is the required probability.

Step-by-step 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:

$z_{score} = \displaystyle\frac{x-\mu}{\sigma}$

P(x < 30,000)

$P( x < 30000) = P( z < \displaystyle\frac{30000 - 26000}{2000}) = P(z<2)$

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

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

Thus, 0.9772 is the required probability.

Answer 2

Answer:

P(X < 30,000) = 0.97725

Step-by-step 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 = $\frac{X-\mu}{\sigma}$ ~ 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( $\frac{X-\mu}{\sigma}$ < $\frac{30000-26000}{2000}$ ) = P(Z < 2) = 0.97725

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