Question

The data (1.5.8.5.1) represent a random sample of the number of days absent from school for five students at Monta Vista High. Find the
mean and standard deviation of the data.
Select one
a. mean = 4; standard deviation = 2.68
b.mean = 4: standard deviation = 7.2
c. mean = 20; standard deviation = 7.6
d. mean = 4.4; standard deviation = 2.76

Answer 1

Answer:

$\bar X = \frac{1+5+8+5+1}{5}= \frac{20}{5}= 4$

$\sigma= \sqrt{\frac{(1-4)^2 +(5-4)^2 +(8-4)^2 +(5-4)^2 +(1-4)^2}{5}} =\sqrt{\frac{36}{5}}= 2.68$

And based on this the best answer would be:

a. mean = 4; standard deviation = 2.68

Step-by-step explanation:

For this case we have the following data given:

1,5,8,5,1

We can find the sample mean with the following formula:

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

And replacing we got:

$\bar X = \frac{1+5+8+5+1}{5}= \frac{20}{5}= 4$

And for the deviation (assuming that the correct approximation is the deviation for a population) we can calculate the deviation with the following formula:

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

And replacing we got:

$\sigma= \sqrt{\frac{(1-4)^2 +(5-4)^2 +(8-4)^2 +(5-4)^2 +(1-4)^2}{5}} =\sqrt{\frac{36}{5}}= 2.68$

And based on this the best answer would be:

a. mean = 4; standard deviation = 2.68