Question

Consider a data set containing the following values: 80, 94, 93, 106, 91, 90, 96, 99 The mean of the preceding values is 93.625. The deviations from the mean have been calculated as follows: –13.625, 0.375, –0.625, 12.375, –2.625, –3.625, 2.375, 5.375 (a) If this is sample data, the sample variance is ___ and the sample standard deviation is ___. (b) If this is population data, the population variance is ___ and the population standard deviation is ___. (c) Suppose the smallest value of 80 in the data was misrecorded as 8. If you were to recalculate the variance and standard deviation with the 8 instead of the 80, your new values for the variance and standard deviation would be ___ and ___.

Answer 1

Answer:

a) $s^2 = \frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}$

And if we replace we got:

$s^2 =\frac{(-13.625)^2 +(0.375)^2 +(-0.625)^2 +(12.375)^2 +(-2.625)^2 +(-3.625)^2 + (2.375)^2 +(5.375)^2}{8-1} = 56.268$

And the sample deviation is just the square root of the variance:

$s = \sqrt{56.268}= 7.501$

b) $\sigma^2 = \frac{\sum_{i=1}^n (X_i -\bar X)^2}{N}$

And if we replace we got:

$\sigma^2 =\frac{(-13.625)^2 +(0.375)^2 +(-0.625)^2 +(12.375)^2 +(-2.625)^2 +(-3.625)^2 + (2.375)^2 +(5.375)^2}{8} = 49.234$

And the deviation would be:

$\sigma = \sqrt{49.234}= 7.018$

c) 8, 94, 93, 106, 91, 90, 96, 99

For this case the new mean would be:

$\bar X = \frac{\sum_{i=1}^n X_i}{n}= 84.625$

The new sample variance would be:

$s^2 =\frac{(-76.625)^2 +(9.375)^2 +(8.375)^2 +(21.375)^2 +(6.375)^2 +(5.375)^2 + (11.375)^2 +(14.375)^2}{8-1} = 984.554$

And the new deviation would be:

$s = \sqrt{984.554}=31.378$

Step-by-step explanation:

For this case we have the following data given:

80, 94, 93, 106, 91, 90, 96, 99

The mean calculated from the following formula is:

$\bar X = \frac{\sum_{i=1}^n X_i}{n}= 93.625$

And the deviations from the mean of the data are:

–13.625, 0.375, –0.625, 12.375, –2.625, –3.625, 2.375, 5.375

Part a

The sample variance is given by the following formula:

$s^2 = \frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}$

And if we replace we got:

$s^2 =\frac{(-13.625)^2 +(0.375)^2 +(-0.625)^2 +(12.375)^2 +(-2.625)^2 +(-3.625)^2 + (2.375)^2 +(5.375)^2}{8-1} = 56.268$

And the sample deviation is just the square root of the variance:

$s = \sqrt{56.268}= 7.501$

Part b

Is the data given comes from a population data then the variance is calculaded as:

$\sigma^2 = \frac{\sum_{i=1}^n (X_i -\bar X)^2}{N}$

And if we replace we got:

$\sigma^2 =\frac{(-13.625)^2 +(0.375)^2 +(-0.625)^2 +(12.375)^2 +(-2.625)^2 +(-3.625)^2 + (2.375)^2 +(5.375)^2}{8} = 49.234$

And the deviation would be:

$\sigma = \sqrt{49.234}= 7.018$

Part c

For this case we have this new data:

8, 94, 93, 106, 91, 90, 96, 99

For this case the new mean would be:

$\bar X = \frac{\sum_{i=1}^n X_i}{n}= 84.625$

The new sample variance would be:

$s^2 =\frac{(-76.625)^2 +(9.375)^2 +(8.375)^2 +(21.375)^2 +(6.375)^2 +(5.375)^2 + (11.375)^2 +(14.375)^2}{8-1} = 984.554$

And the new deviation would be:

$s = \sqrt{984.554}=31.378$