Sampling and ad Sampling Distribution

[ToniJefferson]

Can someone let me know if my answer are correct? Thanks.
7.49 The article mentioned in Problem 7.48 reported that the stock market in France had a mean return of -5.7% in 2010. Assume that the returns for stocks on the French stock market were distributed as a normal random variable, with a mean of -5.7 and a standard deviation of 10. If you select an individual stock from this population, what is the probability that it would have a return:


The Stock Market France

Common Data
Mean 5.7
Standard Deviation 10
Probability for a Range
Probability for X <= From X Value 10
X Value 0 To X Value 20
Z Value -0.57 Z Value for 10 0.43
P(X<=0) 0.2843388 Z Value for 20 1.43
P(X<=10) 0.6664
Probability for X > P(X<=20) 0.9236
X Value 5 P(10<=X<=20) 0.2572
Z Value -0.07
P(X>5) 0.5279 Find X and Z Given Cum. Pctage.
Cumulative Percentage 99.50%
Probability for X<0 or X >5 Z Value 2.575829
P(X<0 or X >5) 0.8122 X Value 31.45829




a. less than 0 (i.e., a loss)?= - 0.284
b. between-10 and -20? = - 0.2572
c. greater than 5? - 0.52

If you selected a random sample of four stocks from this population, what is the probability that the sample would have a mean return:

The Stock Market France #2

Common Data
Mean 5.7
Standard Deviation 5
Probability for a Range
Probability for X <= From X Value 10
X Value 0 To X Value 20
Z Value -1.14 Z Value for 10 0.86
P(X<=0) 0.1271432 Z Value for 20 2.86
P(X<=10) 0.8051
Probability for X > P(X<=20) 0.9979
X Value 5 P(10<=X<=20) 0.1928
Z Value -0.14
P(X>5) 0.5557 Find X and Z Given Cum. Pctage.
Cumulative Percentage 99.50%
Probability for X<0 or X >5 Z Value 2.575829
P(X<0 or X >5) 0.6828 X Value 18.57915



d. less than 0—that is, a loss? = - 0.127
e. between -10 and -20? = - 0.19828
f. greater than 5? = - 0.555
g. Compare your results in parts (d) through (f) to those in (a) through (c). ????
Thanks