SOLUTION: 5. A mutual fund manager is reviewing her account status and finds that the account population is normally distributed with a mean of $60,000 and a standard deviation of $6,000.

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Question 312128: 5. A mutual fund manager is reviewing her account status and finds that the account population is normally distributed with a mean of $60,000 and a standard deviation of $6,000.
a. If she selects an account at random, what is the probability the account will have a value between $60,000 and $69,000? _______
b. If she selects an account at random, what is the probability the account will have a value between $51,000 and $66,000? ________
c. If she selects an account at random, what is the probability the account will have a value less than $55,000? ________
d. If she selects an account at random, what is the probability the account will have a value greater than $72,000? ________
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
a) Find the z score for 60000 and 69000
z%2860000%29=%2860000-60000%29%2F6000=0
z%2869000%29=%2869000-60000%29%2F6000=1.5
P%280%29=0.50000
P%281.5%29=0.93319
P(60K< x <69K)=P(1.5)-P(0)=0.93319-0.5=highlight_green%280.43319%29
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b) Find the z score for 55000
z%2851000%29=%2851000-60000%29%2F6000=-1.5
z%2866000%29=%2866000-60000%29%2F6000=1
P%28-1.5%29=0.066807
P%281%29=0.841345
P(51K< x < 66K)=0.841345-0.066807=highlight_green%280.774538%29
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c) Find the z score for 55000
z%2855000%29=%2855000-60000%29%2F6000=-0.833
P%28-0.8333%29=0.202328
P(x <55K)=highlight_green%280.202328%29
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d) Find the z score for 72000
z%2872000%29=%2872000-60000%29%2F6000=2
P%282%29=0.97725
P(x <72K)=0.97725
P(x >72K)=1-P(x <72K)=1-0.97725=highlight_green%280.02275%29
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