Question

deviation for this population is $40 and that it is normally distbuted Arandom sample of 15 men who According to a research institution, men spontan average of $137 24 on Valentine's Day gifts in 2009. Assume the stand celebrate Valentine's Day was selected Complete parts a through e a Calculate the standard error of the mean ; - 1033 (Round to two decimal places as needed) b. What is the probability that the sample mean will be less than $1257 POR$12) - 1180 (Round to four decimal places as needed) What is the probability that the sample mean will be more than $1402 PG>$140) - (Round to four decimal places as needed)

Answer 1

a) Standard error = 40/sqrt(15)=40/3.87=10.33

b) P ( X<125 )=P ( X−μ<125−137.24 )=P ((X−μ)/σ<(125−137.24)/10.33)
Since (x−μ)/σ=Z and (125−137.24)/10.33=−1.18 we have:
P (X<125)=P (Z<−1.18)
Use the standard normal table to conclude that:
P (Z<−1.18)=0.1190

c) P ( X>140 )=P ( X−μ>140−137.24 )=P ((X−μ)/σ>(140−137.24)/10.33)
Since Z=(x−μ)/σ and (140−137.24)/10.33=0.27 we have:
P ( X>140 )=P ( Z>0.27 )
Use the standard normal table to conclude that:
P (Z>0.27)=0.3936