Respuesta :
Answer:
c. standard deviation, median, range
Step-by-step explanation:
The standard deviation without the Bessel's correct is defined as:
[tex] s= \sqrt{\frac{\sum_{i=1}^n (x_i -\bar x)^2}{n}[/tex]
And if we find the expected value for s we got:
[tex] E(s^2) = \frac{1}{n} \sum_{i=1}^n E(x_i -\bar x)^2 [/tex]
[tex] E(s^2)= \frac{1}{n} E[\sum_{i=1}^n ((x_i -\mu)-(\bar x -\mu)^2)][/tex]
We have this:
[tex] E(\sum_{i=1}^n(x_i-\mu)^2) =n\sigma^2[/tex]
[tex]E[\sum_{i=1}^n (x_i -\mu)(\bar x -\mu)]= \sigma^2[/tex]
[tex]E[\sum_{i=1}^n (\bar x -\mu)^2]=\sigma^2[/tex]
[tex]E(s^2)=\frac{1}{n} (n\sigma^2 -2\sigma^2 +\sigma^2)[/tex]
[tex]E(s^2)=\frac{n-1}{n}\sigma^2[/tex]
as we can see the sample variance is a biased estimator since:
[tex]E(s^2)\neq \sigma^2[/tex]
And we see that the standard deviation is biased, since:
[tex] E(s) = \sqrt{\frac{n-1}{n}} \sigma[/tex]
because [tex]E(s)\neq \sigma[/tex]
The mean is not biased for this case option a is FALSE.
The proportion is not biased for this reason option d is FALSE
The range can be considered as biased since we don't have info to conclude that the range follows a distirbution in specific.
The sample median "is an unbiased estimator of the population median when the population is normal. However, for a general population it is not true that the sample median is an unbiased estimator of the population median".
And for this reason the best option is c.
