probability question need help

Question

Casey is conducting market research on the amount typically spent on the purchase of a first vehicle. Casey has determined that the average cost of a first vehicle is $7800, and that the distribution of costs is normally distributed with a standard deviation of $520. 

a. Find the probability that a single randomly selected first-time vehicle purchase is more than $6950. (Round to 4 decimal places.)

b. The cost of the middle 30% of first-time vehicle purchases falls between (round to the nearest dollar):

$ to $

c. If a random sample of 50 first-time vehicle purchasers is selected, what is the probability that the sample average amount spent is between $7760 and $7950? (Round to 4 decimal places.)

 

Answer 1

Mean is 7800 SD is 520

x is cost

a) P(x > 6950)

translate inequality to z where z = (x - mean)/SD

so P(x > 6950) = P(z > (6950 - 7800)/520) = P(z > -1.63) = P(z < 1.63) = 0.9484

b) middle 30% goes from 15% below and above 50th percentile, so from 35th percentile to 65th percentile.

z = (x - mean)/SD. Find z associated with 65th percentile ~ 0.385

0.385 = (x - 7800)/520

x = 520 * 0.385 + 7800 = 8000

that is $200 above mean.

since normal distribution is symmetric, 35th percentile will be $200 below mean or $7600

range is $7600 to $8000

c) P(7760 < xbar < 7950)

translate inequality to z where z = (xbar - mean)/(SD/sqrt(n)), where n = 50

P(7760 < xbar < 7950) = ( (7760 - 7800)/(520/sqrt(50)) < z < (7950 - 7800)/(520/sqrt(50)) =

P(-0.54 < z < 2.03) = P(z < 2.03) - P(z < -0.54) = 0.9788 - 0.2946 = 0.6842