Question

The selling price of a new car is normally distributed with an average of $25260 and a variance of $13838400.00.

a) What proportion of new cars will sell for less than $18870?
probability =

b) Assuming a normal distribution, within what selling prices will the middle 91% fall?
lower =  , upper =

Answer 1

Solution :

Given that ,

mean = $\mu$ = 25260

variance =  $\sigma$² = 13838400.00

standard deviation = $\sigma$ = $\sqrt{}$ $\sigma$² = $\sqrt{}$ 13838400.00 = 3720

a) P(x < 18870) = P[(x - $\mu$ ) / $\sigma$ < (18870 - 25260) /3720 ]

= P(z < -1.72)

Using z table,

= 0.0427

b) Using standard normal table,

P( -z < Z < z) = 91%

= P(Z < z) - P(Z <-z ) = 0.91

= 2P(Z < z) - 1 = 0.91

= 2P(Z < z) = 1 + 0.91

= P(Z < z) = 1.91 / 2

= P(Z < z) = 0.955

= P(Z < 1.70) = 0.955

= z  ± 1.70

Using z-score formula,

x = z * $\sigma$ + $\mu$

x = -1.70 * 3720 + 25260

x = 18936

Using z-score formula,

x = z * $\sigma$ + $\mu$

x = 1.70 * 3720 + 25260

x = 31584

lower = 18936

upper = 31584

Answer 2

SOLUTION :

Average selling price of car = mean price = 25260 ($)

Variance = 13838400.00 ($^2) 

=> SD = sqrt(variance) = sqrt(13838400.00) = 3720 ($)

a. 

P(x < 18870) 

= P(z < ((18870 - 25260)/3720) 

= P(z < - 1.71)

= 0.0436 (from cumulative ND table). (ANSWER).

b.

Let +/- z be the z-scores interval for 91% area .

=> P( between - z and z) = 0.91

=> P(between z and 0) = 0.455

From ND table, z = 1.695 

So, interval for 91% area is - 1.695 to 1.695 (z-scores) 

Lower limit :

= - z 

= - 1.695

=> x = - z * SD + mean

=> x = - 1.695 * 3720 + 25260

 => x = lower limit of car price = 18954.60 (ANSWER).

Upper limit :

= z 

= 1.695

=> x = z * SD + mean

=> x = 1.695 * 3720 + 25260

=> x = upper limit of car price = 31565.4 (ANSWER).