Suppose that the medium coffees sold by a certain coffee shop have a mean volume of 16 ounces and a standard deviation of 0.4 ounces. What is the probability that a random sample of 100 medium coffees...

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Suppose that the medium coffees sold by a certain coffee shop have a mean volume of 16 ounces and a standard deviation of 0.4 ounces. What is the probability that a random sample of 100 medium coffees sold by that shop has a total volume of less than 1603 ounces?
1 The total evaporation amount of a growing season in Minneapolis is approximately nor-mally distributed with a standard deviaion of 4.00 inches. Given that the most recent 16 growing seasons have an average evaporation amount of 34.50 inches, find a 95% con-fidence interval for the population mean of the growing season evaporation amount in Minneapolis.
3
Historically, the precipitation of July in Twin Cities (in inches) approximately follows a normal distribution with a mean p = 3.57 and a variance a2 = 6.14. Would you still consider a2 = 6.14 a valid value of the variance given that the sample variance for the precipitation of July in Twin Cities during the most recent 30 years is 32 — 11.08?
4 Let X denote the number of days with snowfall of at least 1 inch in a winter. During the past 121 winters, Twin Cities, MN has observed a mean of 7Te = 14.4 days and a sample standard deviation of src = 5.27 days; during the past 61 winters, Eau Claire, WI has observed a mean of .1-kr = 14.9 days and a sample standard deviation of SKC = 5.20 days. Assume that the data from Twin Cities and Eau Claire are independent. Construct a 90% confidence interval for ,,Eza,s, ,, where arc and CEC are the population standard deviations for Twin Cities and Eau blaire, respectively.

Answered Same DayDec 23, 2021

Answer To: Suppose that the medium coffees sold by a certain coffee shop have a mean volume of 16 ounces and a...

Robert answered on Dec 23 2021
117 Votes
Q1
Sample mean=1603/100 =16.03
P( sample mean <16.03) =P( z <(16.03-1600)*(√100)/.4) = P ( z < 0.7
5) = 0.77337
Q2
The interval will be
Sample mean ± 1.96*SE
Where SE= 4 /√16 =1
The interval is 34.5 ± 1.96 =32.54 to 36.46
Q3
The test is a chi square test for variance
Ho: var = 6.14
H0: var ≠ 6.14
The test statistic is (n-1) sample var/population var= 29*11.08/6.14 = 52.33
The critical chi square value is with 29 degrees freedom is 45.7 and 95% confidence level
As...
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