According to a research institution, men spent an average of $136.94 on Valentin

Question

According to a research institution, men spent an average of $136.94 on Valentine's Day gifts in 2009. Assume the standard deviation for this population is $30 and that it is normally distributed. A random sample of 10 men who celebrate Valentine's Day was selected. Complete parts a through e. a. Calculate the standard error of the mean. sigma_x = (Round to two decimal places as needed.) b. What is the probability that the sample mean will be less than $130? P(x $140) = (Round to four decimal places as needed.) d. What is the probability that the sample mean will be between $115 and $155? P($115 lessthanorequalto x lessthanorequalto $166) = (Round to four decimal places as needed.) e. Identify symmetrical interval that includes 95% of the sample means if the true population mean is $135.94. $ lessthanorequalto x lessthanorequalto $ (Round to the nearest dollar as needed.)

Explanation / Answer

given mean =136.94

standard deviation =30

n=10

a) standard error formulea is standard deviation /sqrtn =30/sqrt10=9.4868

b) p(x<=130)=

z= (x^-mu )/s igma = (130-135.94)/30=-0.198

p(z<=-0.198) = 0.4247 from standard z table the value is taken

c)p(x>=140)

z=(x^-mu )/s igma = (140-135.94)/30 =0.1353

p(z>=0.1353) =1-p(x<=0.1353) = 1-0.5517 = 0.4483

d) p(115<=x<=155)

z1= (155-135.94) /30 =0.6353

z2= (115-135.94)/30=-0.698

p(115<=x<=155) = p(z<=0.6353<x<=-0.698) = 0.7357-0.2451=0.4906

e) 95%ci means critical value Za=1.96

x+-Za *sd/sqrtn

135+-1.96*9.4868

135+18.59 <x<135-18.59

153.59<x<116.41

rounding to nearest decimal

154 <x<116

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