Question

A sample of 12 measurements has a mean of 37 and a standard deviation of 4.25. Suppose that the sample is enlarged to 14 measurements, by including two additional measurements having a common value of 37 each.

A. Find the mean of the sample of 14 measurements.

Mean = 37

How I found it: 12 (the original sample) * 37 (mean) = 444 + (2*37, the two new scores) = 518/14 (the new sample size) = 37

WHAT I NEED HELP WITH (i.e. how can i calculate it?):

B. Find the standard deviation of the sample 14 measurements.

Standard Deviation =

Answer 1

To find the standard deviation of the sample of 14 measurements, you can use the formula for population variance (since you have all the data points):

Variance = (Σ(x - μ)^2) / n

where μ is the mean, which is 37, and n is the number of measurements, which is 14 for the enlarged sample.

But before that, let's find the sum of squared deviations (SSD) from the mean for the original 12 measurements:

SSD_12 = (standard deviation_12)^2 * (sample size_12)
SSD_12 = (4.25^2) * 12
SSD_12 = 216.75

Now let's find the SSD for the enlarged sample of 14 measurements:

SSD_14 = SSD_12 + (37 - 37)^2 + (37 - 37)^2
SSD_14 = 216.75 + 0 + 0
SSD_14 = 216.75

Now find the variance for the sample of 14 measurements:

Variance_14 = SSD_14 / 14
Variance_14 = 216.75 / 14
Variance_14 ≈ 15.48

Lastly, to find the standard deviation, take the square root of the variance:

Standard Deviation_14 = √Variance_14
Standard Deviation_14 = √15.48
Standard Deviation_14 ≈ 3.93

So, the standard deviation of the sample of 14 measurements is approximately 3.93.