The manager of The Cheesecake Factory in Boston reports that on six randomly selected weekdays, the number of customers served w
as 120, 130, 100, 205, 185, and 220. She believes that the number of customers served on weekdays follows a normal distribution. Construct the 90% confidence interval for the average number of customers served on weekdays. (You may find it useful to reference the t table. Round intermediate calculations to at least 4 decimal places. Round "sample mean" and "sample standard deviation" to 2 decimal places and "t" value to 3 decimal places and final answers to 2 decimal places.)
1 answer:
Answer:
143.31 < µ < 176.69
Step-by-step explanation:
The manager wants a 90% confidence interval based on the data. We are given 6 weekdays of data. We have to do 3 things:
1. Calculate the mean number of customers
2. Calculate the standard deviation of the number of customers
3. Construct the confidence interval
*For #1 and #2, see attached photo 1 for the calculations...
The sample mean is: 160
The sample standard deviation is: 20.29
The t-value we need is found using the t-distribution chart, with 5 degrees of freedom. Our t-value is: 2.015
*See attached photo 2 for construction of the confidence interval
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Answer:
Team A.
Step-by-step explanation:
We know that Team A had 10 of its 12 members present.
We can write this as the following fraction:
We can convert this to a decimal by dividing the two numbers. Use a calculator:
And we can convert this to a percent by moving the decimal too places to the right and by adding a percent sign:
So, about 83.33% of members on Team A was present.
We know that Team B had 13 of its 15 members present.
We can write this as the following fraction:
Again, we can convert this to a decimal by dividing the two numbers:
Change this to a percent:
So, Team A had a participation rate of 83.33% while Team B had a participation rate of 86.67%.
So, the team that had the lesser percent of the team present at the tournament is Team A.
And we're done!