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.

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) (Show Source): You can put this solution on YOUR website!
a) Find the z score for 60000 and 69000

P(60K< x <69K)=P(1.5)-P(0)=0.93319-0.5=
.
.
.
b) Find the z score for 55000

P(51K< x < 66K)=0.841345-0.066807=
.
.
.
c) Find the z score for 55000

P(x <55K)=
.
.
.
d) Find the z score for 72000

P(x <72K)=
P(x >72K)=1-P(x <72K)=1-0.97725=
.
.
.
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