Academic Integrity: tutoring, explanations, and feedback — we don’t complete graded work or submit on a student’s behalf.

A study of pre-owned Play Station 4 prices is done, and has been determined that

ID: 3173664 • Letter: A

Question

A study of pre-owned Play Station 4 prices is done, and has been determined that they are normally distributed with mu = $105 and sigma = $8.50 Determine the percentage of Play Station 4's that are between $95 and S115 Determine the probability that a randomly selected Play Station 4 cost the $120 A consumer report in Gamer Magazine recommends not purchasing used Play Station 4's that are in the highest 2% of the normal price range. Using the mean, and standard deviation from above, what is the most that a person should speed in a used Play Station 4? A small Game Stop store is ordering a group of 5 used Play Station 4s. What is the probability that the mean price or the group is above $120? (PS) The same Game Stop store is purchasing a second batch of 6 used Play Stations. What is the probability that the total price will be more than $600?

Explanation / Answer

a)P(95<X<115)=P((95-105)/8.5<Z<(115-105)/8.5)=P(-1.1765<Z<1.1765)=0.8803-0.1197=0.7606

b)P(X>120)=1-P(X<120)=1-P(Z<(120-105)/8.5)=1-0.9612=0.0388

c)for highest 2%, at 98 percentile, z=2.0537

hence corresponding value =mean +z*std deviation =122.4569

d)for mean price, std error of mean =std deviation/(n)1/2=3.80

hence P(X>120)=1-P(X<120)=1-P(Z<(120-105)/3.8)=1-P(Z<3.946)=1-0.99996=0.00004

e) for 6 playstation expected prcie =6*105=630

and std deviation =8.5*(6)1/2=20.82

hence P(X>600) =1-P(X<600)=1-P(Z<(600-630)/20.82)=1-P(Z<-1.4409)=1-0.0748=0.9252