Aresearch firm conducted a survey to determine the mean amount americans spend on coffee during a week. they found the distribution of weekly spending followed the normal distribution with a population standard deviation of $5. a sample of 64 americans revealed that x¯¯¯=$20 . what is the 95% confidence interval estimate of μ?

95% confidence Interval would be between $18.775 and $20.225.

Step-by-step explanation:

Confidence Interval can be calculated using M±ME where

And margin of error (ME) around the mean calculated as

ME= where

Using the numbers, we get:

ME= =1.225

Then 95% confidence Interval would be 20±1.225 or between $18.775 and $20.225

so what is the question

step-by-step explanation:

answer: first option i.e. 'a' is correct.

step-by-step explanation:

since we have given that

number of minutes to wrap a small gift box = 4 minutes

we need to find the number of gifts to be wrapped in 5 minutes

since there is a direct variation, as the number of minutes will increase with   number of gifts to be wrapped.

so, let the number of gifts to be wrapped in 5 minutes be x.

so, it becomes,

hence, first option 'a'   is correct.