Anew york restaurant would like to estimate the average amount customers spend for dinner. a sample of size 49 customers resulted an average of $32.60. assume the population standard deviation is $5.what is the point estimation for the mean of the population? at 90% level of confidence, what is the margin of error? what is the 90% confidence level for the mean of the population?

Point of estimation = 32.60

Margin of error = 1.175.

95% confidence interval would be (81.95,84.45).

Step-by-step explanation:

Since we have given that

n = 49

Average = $32.60

Standard deviation = $5

We need to find:

1) Point of estimation for the mean of the population:

2) At 90% level of confidence, z = 1.645

Margin of error would be

So, Margin of error = 1.175.

3) At 90% level of confidence, interval would be

Hence, 90% confidence interval would be (31.43,33.76).

d.192

step-by-step explanation:

12*16=192

a.   since the profit made in 2012 is equal to the profit made in 2014, the profit is symmetric about the line y = 120,000.

step-by-step explanation: