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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?

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  • $\begingroup$ 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
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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

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