Question

Hey!! CAN SOMEONE PLEASE ANSWER FOR ME QUICK THANKYOU!

One college class had a total of 80 students. The average score for the class on the last exam was 82.6 with a standard deviation of 5.3. A random sample of 35 students was selected. a. Calculate the standard error of the mean. b. What is the probability that the sample mean will be less than 84? c. What is the probability that the sample mean will be more than 83? d. What is the probability that the sample mean will be between 81.5 and 83.5? a. The standard error of the mean is - (Round to two decimal places as needed.) . b. The probability that the sample mean will be less than 84 is (Round to four decimal places as needed.) : C. The probability that the sample mean will be more than 83 is (Round to four decimal places as needed.) | d. The probability that the sample mean will be between 81.5 and 83.5 is (Round to four decimal places as needed.)

Answer 1

Mean = $\mu$ = 82.6

Standard deviation = $\sigma$ = 5.3

Sample size = n = 35

a)

The standard error of the mean is

$\frac{\sigma}{\sqrt{n}}=\frac{5.3}{\sqrt{35}}=0.90$

b)

We have to find P( $\bar{x}$ < 84)

For finding this probability we have to find z score.

UMO

$z =\frac{84-82.6 }{5.3 /\sqrt{35}}=\frac{1.4}{0.90}=1.56$

That is we have to find P(Z < 1.56)

P(Z < 1.56) = 0.9409

( From z table)

c)

We have to find P( $\bar{x}$ > 83)

For finding this probability we have to find z score.

UMO

$z =\frac{83-82.6 }{5.3 /\sqrt{35}}=\frac{0.4}{0.90}=0.46$

That is we have to find P(Z > 0.46)

P(Z > 0.46) = 1 - P(Z < 0.46) =1 - 0.6724 =  0.3276

( From z table)

d)

We have to find P( 81.5 < $\bar{x}$ < 83.5)

For finding this probability we have to find z score.

UMO

$z =\frac{81.5-82.6 }{5.3 /\sqrt{35}}=\frac{0.90}{0.90}=-1.23$

$z =\frac{83.5-82.6 }{5.3 /\sqrt{35}}=\frac{0.90}{0.90}=1$

That is we have to find P( - 1.23 < Z < 1)

P( - 1.23 < Z < 1) = P(Z < 1) - P(Z < - 1.23) = 0.8413 - 0.1097 =  0.7316

( From z table)