Apopulation of 1,000 students spends an average of $10.25 a day on dinner. the standard deviation of the expenditure is $3. a simple random sample of 64 students is taken.

a. what are the expected value, standard deviation, and shape of the sampling distribution of the sample mean?

b. what is the probability that these 64 students will spend a combined total of more than $715.21?

c. what is the probability that these 64 students will spend a combined total between $703.59 and $728.45?

Step-by-step explanation:

Given that a population of 1000 students spends an average of $10.25 a day on dinner.

The standard deviation of the expenditure is $3.

X , the amount spend is N(10.25, 3)

n =64

a) Mean of sample mean = 10.25

Std devition of sample mean = std deviation /sq rt of n = 0.375

The shape would be bell shaped symmetrical about the mean and unimodal.

b) Prob that these 64 students will spend a combined total of more than $715.21

=

c) the probability that these 64 students will spend a combined total between $703.59 and $728.45

= x bar lies between = 10.9936 and 11.38203

=

283 square feet

step-by-step explanation:

good luck

the answer would be c:   ax + b = ax + c

step-by-step explanation: