5. According to us. News & word Report\'s publication America\'s Best Colleges,

Question

5. According to us. News & word Report's publication America's Best Colleges, the average cost to attend the University of Southern Calfornia (SC) aner deducting grants based on need is $26,450. Assume the population standard deviation is $6,900. Suppose that a random sample of 40 USC students will be taken from this population. Use z-table a. What is the value of the standard error of the mean? to nearest whole number) 10. 0 b. What is the probability that the sample mean will be more than 26,450 (to 2 decimals) c. What is the probablity that the sample mean will be within $750 of the population mean? (to 4 decimals) d. How would the probability in part (c) change if the sample size were increased to 160? to 4 decimals)

Explanation / Answer

mean= $ 26450 , standard deviation= $6900 and n= 40

a) standard error= 6900/sqrt(40)= 6900/6.324555= $1091

b) Since =26450 and =1091 we have:

P ( X>26450 )=P ( X>2645026450 )=P ( (X)/>(2645026450)/1091)

Since Z=(x)/ and (2645026450)/1091=0 we have:

P ( X>26450 )=P ( Z>0 )

Use the standard normal table to conclude that:

P (Z>0)=0.50

c) Since =26450 and =1091 we have:

P ( X<750 )=P ( X<75026450 )=P ((X)/<(75026450)/1091)

Since (x)/=Z and (75026450)/1091=23.56 we have:

P (X<750)=P (Z<23.56)

Use the standard normal table to conclude that:

P (Z<23.56)=0.0000

d) standard error= 6900/sqrt(160)= 546

Since =26450 and =546 we have:

P ( X<750 )=P ( X<75026450 )=P ((X)/<(75026450)/546)

Since (x)/=Z and (75026450)/546=47.07 we have:

P (X<750)=P (Z<47.07)

Use the standard normal table to conclude that:

P (Z<47.07)=0.0000

Related Questions

Navigate

• Browse (All)
• Browse 5
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.