Sampling distribution for the mean of a large sample of salaries

Question

Hello I am trying to solve this question. But, I can not find any approach, We know that avg is 600 and the question is asking for 500? and also if we sum the sd and avg we can not come near to the 1000£ on the second part. Could you please help me to solve this question?

Average income of nurses is 600£ and the standard deviation was 266£ . A sample of 1,000 nurses was randomly chosen to verify if this sample would have a similar mean income as the entire population. a) Probability that the mean income of the nurses sampled is less than 500£ . b) Probability that the mean income of the nurses sampled is greater than 1000£ .

Answer 1

Use the Central Limit Theorem: the sample mean $\bar{X} = \frac{1}{n}\sum_{i=1}^n X_i$ converges to $$ \mathcal{N}(\mu, \sigma^2/n) $$ in distribution, with $n = 1000$, $\mu = 600$ and $\sigma = 266$.

Therefore let $$ f_\bar{X}(x) = \dfrac{1}{\sqrt{2\pi\sigma^2}} \cdot e^{\frac{-(x-\mu)^2}{2\sigma^2}} $$ be the p.d.f. of the normal distribution given above, (a) is $$ \mathbb{P}[\bar{X} \leq 500] = \int_{-\infty}^{500} f_\bar{X}(x) dx $$

and (b) is simply $$ \mathbb{P}[\bar{X} \geq 1000] = 1 - \mathbb{P}[\bar{X} \leq 1000] = 1 - \int_{-\infty}^{1000} f_\bar{X}(x) dx $$

Answer 2

Your idea of looking at the number of standard deviations $\sigma$ from the mean $\mu$ is a reasonable approach.

If the distribution of incomes is $\mathsf{Norm}(\mu=600, \sigma = 266),$ then $\bar X_{1000} \sim \mathsf{Norm}(600, 266/\sqrt{1000}),$ where $SD(\bar X_{1000}) = 266/\sqrt{1000} = 8.4117.$

By the Central Limit Theorem the distribution for $\bar X_{1000}$ is very nearly normal even if the population of nurse incomes is not normally distributed.

In (a) you seek $P(\bar X_{1000} < 500) \approx 0.$ This is asking for the probability in a normal distribution that lies to the left of about eleven standard deviations below its mean, so you should not be surprised by the nearly zero probability.

For more precision in such computations, you can 'standardize' and look at printed standard normal CDF tables use software (calculator or computer). Use whatever method is discussed in your course text or lectures. I used R statistical software to get the value stated above.

pnorm(500, 600, 266/sqrt(1000))
[1] 6.810586e-33

(b) Similar:

1 - pnorm(1000, 600, 266/sqrt(1000))
[1] 0

Note: Maybe a more user-friendly initial problem of this kind would have asked for $P(\bar X_{1000} < 590) = 0.1173$ and $P(\bar X_{1000} > 615) = 0.0373.$