The distribution of weights of 1000 students is normal with a mean of 55 kg and a variance of 25 kg. 100 random samples of size 16 are taken from this population. Determine the following:

i) The mean and standard deviation of the sampling distribution?

How do i find the standard deviation?

  • $\begingroup$ Weights measured in kilogram cannot follow a normal distribution, as it would predict the existence of negative weights. Maybe they are lognormal. $\endgroup$ – Björn Friedrich Nov 20 '16 at 15:05
  • $\begingroup$ @BjörnFriedrich. Technically, you're right. But in applications, normal can still be a useful approximation if the mean (here 55) is more than three or four standard deviations (here 5) above 0. If X∼Norm(μ=55,σ=5), then the implied probability P(X<0)= 1.91066e-28. An event that wouldn't occur in anyone's lifetime. However, in applications weights may not be precisely modeled by normal distributions because they tend to be slightly right-skewed. People more than 2 SDs above the mean wt may (indelicately) be called 'fat'; those more than 2 SDs below the mean wt often (unfortunately) 'dead'. $\endgroup$ – BruceET Nov 20 '16 at 21:49

You are probably studying, or about to study, the Central Limit Theorem. You should find some explanation of this in your text--adjacent to the discussion of the CLT. In particular, a sample mean of observations from a normal population is normally distributed.

If $X_1, X_2, \dots, X_{16}$ are a random sample from $Norm(\mu = 55, \sigma = 5),$ then the sample mean $\bar X \sim Norm(\mu = 55, \sigma = 5/\sqrt{16}) = Norm(5, 5/4).$

Key steps in the derivation of $E(\bar X) = \mu$ are as follows: $$E(\bar X) = E\left(\frac{1}{n}\sum_{i=1}^n X_i\right) = \frac{1}{n}E\left(\sum_{i=1}^n X_i\right) = \frac{1}{n}\sum_{i=1}^n E(X_i) = \frac{1}{n}(n\mu) = \mu.$$ Notice that each of the $n$ terms in the last summation is $\mu.$

Key steps in the derivation of $Var(\bar X) = \sigma^2/n$ are as follows, where the first step uses the identity $Var(aY) = a^2Var(Y):$

$$Var(\bar X) = Var\left(\frac{1}{n}\sum_{i=1}^n X_i\right) = \left(\frac{1}{n}\right)^2 Var\left(\sum_{i=1}^n X_i\right) \\= \frac{1}{n^2}\sum_{i=1}^n Var(X_i) = \frac{1}{n^2}(n\sigma^2) = \sigma^2/n.$$

Then $SD(\bar X) = \sqrt{Var(\bar X)} = \sqrt{\sigma^2/n} = \sigma/\sqrt{n}.$

Example: As a consequence, the probability that any one individual in the population weighs between 50 and 60kg is $P(50 <X_i < 60) \approx .68$ However, the probability that the average weight of 16 individuals lies between 50 and 50kg is $P(50 < \bar X < 60) \approx 1.$

 diff(pnorm(c(50,60), 55, 5))
 ## 0.6826895
 diff(pnorm(c(50,60), 55, 5/4))
 ## 0.9999367

The plot below shows the normal PDF of the population and the normal PDF of $\bar X.$

enter image description here


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.