Question

A sample of 12 measurements has a mean of 38 and a sample standard deviation of 4.25. Suppose that the sample is enlarged to 14 measurements, by including two additional measurements having a common value of 38 each. The sample standard deviation of the 14 measurements is:

Answer 1

Answer:

The sample standard deviation of the 14 measurements is 3.93

Step-by-step explanation:

The standard deviation = √(variance)

The standard deviation is the square root of variance. And variance is an average of the squared deviations from the mean.

Mathematically,

Standard deviation = σ = √[Σ(x - xbar)²/N]

x = each variable

xbar = mean

N = number of variables

For 12 variables,

N = 12

σ = 4.25

xbar = 38

Σ(x - xbar)² = sum of the square of all deviations; let it be equal to D

4.25 = √[Σ(x - xbar)²/12]

4.25 =√(D/12)

4.25² = D/12

D = 216.75.

For 14 measurements,

N = 14

The mean is going to still be 38, because the two new measurements are each 38.

xbar = 38

And the new additions to the sum of deviations, will be (38 - 38)² twice, that is, 0.

Standard deviation for 14 measurements

Standard deviation = σ = √[Σ(x - xbar)²/N]

σ = √(216.75/14) = 3.93