1) A random sample is drawn from a normally distributed population with mean = 3
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Question
1) A random sample is drawn from a normally distributed population with mean = 31 and standard deviation = 1.9. Use Table 1.
Calculate the above probabilities for both sample sizes. (Round intermediate calculations to 4 decimal places, “z” value to 2 decimal places, and final answer to 4 decimal places.)
76?
2) A small hair salon in Denver, Colorado, averages about 25 customers on weekdays with a standard deviation of 8. It is safe to assume that the underlying distribution is normal. In an attempt to increase the number of weekday customers, the manager offers a $5 discount on 6 consecutive weekdays. She reports that her strategy has worked since the sample mean of customers during this 6 weekday period jumps to 31. Use Table 1.
What is the probability to get a sample average of 31 or more customers if the manager had not offered the discount? (Round “z” value to 2 decimal places, and final answer to 4 decimal places.)
1) A random sample is drawn from a normally distributed population with mean = 31 and standard deviation = 1.9. Use Table 1.
Calculate the above probabilities for both sample sizes. (Round intermediate calculations to 4 decimal places, “z” value to 2 decimal places, and final answer to 4 decimal places.)
n Probability 39?76?
2) A small hair salon in Denver, Colorado, averages about 25 customers on weekdays with a standard deviation of 8. It is safe to assume that the underlying distribution is normal. In an attempt to increase the number of weekday customers, the manager offers a $5 discount on 6 consecutive weekdays. She reports that her strategy has worked since the sample mean of customers during this 6 weekday period jumps to 31. Use Table 1.
a.What is the probability to get a sample average of 31 or more customers if the manager had not offered the discount? (Round “z” value to 2 decimal places, and final answer to 4 decimal places.)
ProbabilityExplanation / Answer
for first question Pls provide what probability needs to be answered. pls revert
2) here mean =25
and std deviation =8
hence std error of mean =std deviation/(n)1/2 where n=6
=3.266
hence P(X>31) =1-P(X<31) =1-P(Z<(31-25)/3.266) =1-P(Z<1.837) =1-0.9669=0.0331
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