Suppose a research firm conducted a survey to determine the mean amount steady s

Question

Suppose a research firm conducted a survey to determine the mean amount steady smokers spend on cigarettes during a week. A sample of 100 steady smokers revealed that the sample mean is $20 and the sample standard deviation is $5. What is the probability that a sample of 100 steady smokers spend between $20.80 and $22.00?

0.2294

0.5328

0.1915

0.1385

0.4967

0.2294

0.5328

0.1915

0.1385

0.4967

enakel picourse assessment id- 42416.18 course ida 1799 Test Information Timed Test This test has atrme of2hours This test wa save and submit automaticaly whenthetme expres Warnings appear when half the time, 5 minutes, 1 minute, and 30 seconds remain, Mutple Attempts This test alows 2 attempts. Thisisatempt number 1 Force Completion once started, this test must be completed in one sning Donotleave the test before clicking Save and submit Remaining Time: 1 hour, 57 minutes, 44 seconds. uestion Completion Status Moving to another question wilsavethis response. Question 3 suppose a research firm conducted a survey to determine the mean amountsteady smokers spend dgaretes durng a week Asample of 100 steady smokers revealed thatthe sample mean is $20 and the sample s deviation is S5 What is the probabnty that asample of 100 steady smokers spend between $20.80 and $22.00? O 0.2294 O 0.5328 O 0.1915 O 0.1385 O 0.4967 Question 3 of 5 Moving to another question wasave this response

Explanation / Answer

n =100 , mean = 20 , std.dviation = 5

By normal distribution formula,

z = (x - mean) / ( s / sqrt(n))

P((20.80 - 20)/(5 / sqrt(100) < z < ( 22- 20) / (5 / sqrt(100)) = (1.6 < z < 4)

Now, we need to find P(1.6 < z < 4)

p(20.80 < x < 22) = p(1.6 < z < 4) = 0.0548

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