Question

Suppose that the average price of a 5 year old used car is $16,230 with a standard deviation of $4,740. Assume that the price follows a normal distribution, find the following

1. The probability of an used car is less than $19,000?
2. The probability of an used car is between 13,000, and 18,000?
3. The probability is between 11,000 and 16,000?
4. The probability is greater than 20,000?
5. I’m looking for an used car in the cheapest 10%. How much am I willing to pay for this car?

Answer 1

Solution :

Given that,

mean = =16,230

standard deviation = =27

A ) P( x < 19,000)

P ( x - / ) < ( 19,000 - 16,230 / 4,740)

P ( z < 2770 / 4,740 )

P ( z < 0.58 )

Using z table

= 0.7190

Probability = 0.7190

B ) P(13000 < x < 18,000)

P( 13,000 - 16,230 / 4,740)< ( x - / ) < ( 18,000 - 16,230 / 4,740)

P ( - 3230 / 4,740 < z < 1770 / 4,740 )

P ( - 0.68 < z < 0.37 )

P ( z < 0.37 ) - P ( z < - 0.68 )

Using z table

= 0.6443 - 0.2483

= 0.3960

Probability = 0.3960

C ) P(13000 < x < 16,000)

P( 11,000 - 16,230 / 4,740)< ( x - / ) < ( 16,000 - 16,230 / 4,740)

P ( - 5230 / 4,740 < z < - 230 / 4,740 )

P ( - 1.10 < z <- 0.05 )

P ( z < - 0.05 ) - P ( z < - 1.10 )

Using z table

= 0.4801 - 0.1357

= 0.3444

Probability = 0.3444

D ) P( x > 20,000)

= 1 - P( x < 20,000)

= 1 - P ( x - / ) < ( 20,000 - 16,230 / 4,740)

= 1 - P ( z < 3770 / 4,740 )

= 1 - P ( z < 0.79 )

Using z table

= 1 - 0.7852

= 0.2148

Probability = 0.2148

E ) Using standard normal table,

P(Z > z) = 10%

1 - P(Z < z) = 0.10

P(Z < z) = 1 - 0.10 = 0.90

P(Z < 1.282) = 0.90

z = 1.28

Using z-score formula,

x = z * +

x = 1.28 * 4740 + 16230

= 22297.2

To pay for this car  $ 22297.2