Answer to Question #328355 in Statistics and Probability for Chanty

Question

The average amount of money that a depositor of the Second National City

Bank has in an account in $5000 with a standard deviation of $650. A

random sample of 36 accounts is taken. What is the probability that the

average amount of money that these 36 depositors have in their accounts

is

a. Between $4800 and $5300?

b. Less than $ 4650 and Greater than $5250

c. Greater than $4900

Answer 1

a.$P(4800<X<5300)=P(\frac{4800-5000}{650/\sqrt{36}}<Z<\frac{5300-5000}{650/\sqrt{36}})=P(-1.85<Z<2.77)=0.9972-0.0322=0.9650$

b. P(X<4650) +(X>5250)=1-P(4650<X<5250)

$P(4650<X<5250)=P(\frac{4650-5000}{650/\sqrt{36}}<Z<\frac{5250-5000}{650/\sqrt{36}})=P(-3.23<Z<2.31)=0.9896-0.0006=0.9890$

P(X<4650) +(X>5250)=1-0.9890=0.0110

c.$P(X>4900)=1-P(X<4900)=1-P(Z<\frac{4900-5000}{650/\sqrt{36}})=1-P(Z<-0.92)=1-0.1788=0.8212$