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Alex_Xolod [135]
3 years ago
5

Alexandra worked 5 hours on Monday, 4.5 hours on Tuesday, and 2.5 hours on Wednesday which is closest to the mean number of hour

s she worked over the three day period
Mathematics
2 answers:
7nadin3 [17]3 years ago
7 0
I think the mean would be 4.5 hours but if this is a multiple choice question can you tell me what the options are
serg [7]3 years ago
3 0
I got the answer 4 .....
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Suppose 60% of American adults believe Martha Stewart is guilty of obstruction of justice and fraud related to insider trading.
Ierofanga [76]

Answer:

Since the sample size is large enough (n>30) and the probability of success is near to 0.5, and we have that n\hat p = 60>10 and n(1-\hat p) = 40>10 we can assume that the distribution for \hat p is normal

The population proportion have the following distribution  

\hat p \sim N(\hat p,\sqrt{\frac{\hat p(1-\hat p)}{n}})  

The mean is given by:

\mu_p = 0.6

\sigma_p = \sqrt{\frac{0.6*(1-0.6)}{100}}=0.04899

So then we can conclude that the best answer would be:

c. approximately Normal, with mean 0.6 and standard error 0.04899

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".  

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

p represent the real population proportion of interest

\hat p represent the estimated proportion for the sample

n is the sample size required (variable of interest)

Solution to the problem

Since the sample size is large enough (n>30) and the probability of success is near to 0.5, and we have that n\hat p = 60>10 and n(1-\hat p) = 40>10 we can assume that the distribution for \hat p is normal

The population proportion have the following distribution  

\hat p \sim N(\hat p,\sqrt{\frac{\hat p(1-\hat p)}{n}})  

The mean is given by:

\mu_p = 0.6

\sigma_p = \sqrt{\frac{0.6*(1-0.6)}{100}}=0.04899

So then we can conclude that the best answer would be:

c. approximately Normal, with mean 0.6 and standard error 0.04899

8 0
3 years ago
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