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ezto.mheducation com Accounting web Slice Gallery Sitee sugaérés Governor in council app eAlgonquin.ca 99 Names of A.teric Onling AC live Nelsonrain- My Home lgin.aro Windows Personnallser les liens Google BMO- Ban of.ine Banking> Exercise 7-46 The mean amount purchased by each customer at Churchill's Grocery Store is $35 with a standard deviation of $9. The population is positively skewed. For a sample of 42 customers, answer the following questions: a. What is the likelihood the sample mean is at least $39? (Round the z-value to 2 decimal places and the final answer to 4 decimal places.) Sample mean b. What is the likelihood the sample mean is greater than $33 but less than $39? (Round the z-value to 2 decimal places and the final answer to 4 decimal places.) Sample mean C. Within what limits will 98% of the sample means occur? (Round the final answers to 2 decimal places.) Sample mean and References eBook & Resources Worksheet Exercise 7-46

Explanation / Answer

a)

b)

=0.0769

c)

for 98% CI ; z =-/+ 2.33

hence 98% CI value =mean -/+ z*std deviation =31.76 and 38.24

( please revert if any answer mismatches due to rounding)

for normal distribution z score =(X-)/ here mean=       = 35.000 std deviation   == 9 sample size       =n= 42 std error=x=/n= 1.3887 probability = P(X>39) = P(Z>2.88)= 1-P(Z<2.88)= 1-0.998= 0.0020

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