SOLUTION: . A youth sports team is trying to determine how many small, medium, and large size jerseys to buy for its players. Small jerseys fit children up to 60 pounds. Medium jerseys fit

Question 1020072: . A youth sports team is trying to determine how many small, medium, and large size jerseys to buy for its players. Small jerseys fit children up to 60 pounds. Medium jerseys fit children 60-75 pounds. Large jerseys fit children 75-110 pounds. The distribution of the players� weights is normally distributed with a mean of 70 pounds and a standard deviation of 10 pounds. If there are 50 kids on the team, how many of each jersey should the team purchase? Be sure to
Answer by mathmate(429) (Show Source): You can put this solution on YOUR website!

Question:
. A youth sports team is trying to determine how many small, medium, and large size jerseys to buy for its players. Small jerseys fit children up to 60 pounds. Medium jerseys fit children 60-75 pounds. Large jerseys fit children 75-110 pounds. The distribution of the players� weights is normally distributed with a mean of 70 pounds and a standard deviation of 10 pounds. If there are 50 kids on the team, how many of each jersey should the team purchase?

Solution:
Nice practical problem!

Find the Z-score for each of the transition weights.
μ=70; σ=10
X=60; Z=(X-μ)/σ=(60-70)/10=-1
X=75; Z=(75-70)/10=0.5
X=110; Z=(110-70)/10=4
Now use the normal distribution table to figure out the following:
For Small, X<=60, P(X<=60)=P(Z<=-1)=0.1587,
number of small jersey's to order = 50*0.1587=7.94=> 8 small jerseys
For medium, 60<=X<=75 => -1<=Z<=0.5 => P(Z<=0.5)=0.6915
number of medium jerseys to order = 50*0.6915-8 = 26.575 => 27 medium jerseys
For large, 75<=X<=110 => P(Z<=4)=.99997
number of large jerseys to order =.99997*50-(8+27)=14.998 => 15 large jeuseys
Check: 8+27+15=50 exactly, i.e. no extras, and hope the the distribution of weights is actually normal.
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