Julie manages a franchise restaurant and is analyzing the tips left at her location. From corporate informational material, she knows the overall
population mean is $6.85 with a standard deviation of $1.25. Julie has a sample of 180 tips for her franchise. By the central limit theorem, which
interval can Julie be 95% certain that the sample mean will fall within?
OA. $6.76 and $6.94
O B. $6.57 and $7.13
OC. $6.66 and $7.04
OD. $6.83 and $6.87

The $6.76 and $6.94 is the interval Julie can be 95% certain that the sample means will fall within option (A) is correct.

What is the standard deviation?

It is defined as the measure of data disbursement, It gives an idea about how much is the data spread out.

[tex]\rm \sigma = \sqrt{\dfrac{ \sum (x_i-X)}{n}[/tex]

σ is the standard deviation

xi is each value from the data set

X is the mean of the data set

n is the number of observations in the data set.

From the central limit theorem:

σ(x) = σ/√n

The standard deviation σ = $1.25

The sample n = 180

σ(x) = 1.25/√180

σ(x) = 1.25/√1801

σ(x) = 0.0931

The lower limit = 6.85 - 0.0931 = 6.756 = 6.76

The upper limit = 6.85 - 0.0931 = 6.943 = 6.94

Thus, the $6.76 and $6.94 is the interval Julie can be 95% certain that the sample means will fall within option (A) is correct.

Learn more about the standard deviation here:

brainly.com/question/12402189

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