Respuesta :
Answer:
mean = 6.15
variance = 4.637
standard deviation = 2.15
Step-by-step explanation:
It is given that 75.4% of women aged 20-24 have never been married.
We are interested in finding the mean, variance and standard deviation for the number who are or who have been married.
So the percentile of women who have been married is given by
[tex]p = 1 - 0.754 \\\\ p = 0.246 \\\\q = 1 - p \\\\q = 1 - 0.246 \\\\q = 0.754[/tex]
The mean is given by
[tex]mean = n\times p[/tex]
Where n is the sample size
[tex]mean = 25\times 0.246 \\\\mean = 6.15[/tex]
Therefore, the mean is 6.15
The variance is given by
[tex]variance = n \times p \times q \\\\variance = 25 \times 0.246 \times 0.754 \\\\variance = 4.637[/tex]
Therefore, the variance is 4.637
The standard deviation is given by
[tex]standard \: deviation = \sqrt{variance} \\\\standard \: deviation = \sqrt{4.637} \\\\standard \: deviation = 2.15[/tex]
Therefore, the standard deviation is 2.15
