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I have a random variable $X_i = \{$ harvest of wheat in quintals per one hectar $\}$ with a uniform distribution $X_i ~ U(16;20)$. The total number of hectars is $1600$. I'm supposed to give in interval within which about 0.99 probability the whole harvest falls, using central limit theorem.

So I calculated that $\eta = 18$ , $\sigma = 4/3$. Then, using the CLT, I have that

$P (\bar{X}_{1600} \le \alpha) = 0.99$

$\alpha = 28907.43$ and this should be my upper bound. My question is, how can I calculate the lower bound of this interval? Thank you very much

1 Answers 1

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Let $ X_i \sim Uniform[16,20] $. Then $ E[X_i] = 18 $ and $ Var[X_i] = \frac{4}{3} $, which means that $ E[\bar{X}_{1600}] = 18 $ and $ Var[\bar{X}_{1600}] = \frac{4}{3(1600)} $.

The question is asking you to find $ a $ and $ b $ such that $ P(a \leq \bar{X}_{1600} \leq b) = 0.99 $.

First, we know that the following equation gives us a two-sided $ (1-\alpha)100\% $ confidence interval for a standard normal distribution:

$$ P(-z_{\alpha/2} \le Z \le z_{\alpha/2}) = 1-\alpha $$

Next, we use the CLT, i.e. $ \frac{\bar{X}_{1600}-E[\bar{X}_{1600}]}{Var[\bar{X}_{1600}]} \rightarrow Z $.

$$ P \bigg(-z_{(0.01/2)} \leq \frac{\bar{X}_{1600}-E[\bar{X}_{1600}]}{Var[\bar{X}_{1600}]} \leq z_{(0.01/2)} \bigg) = 0.99 $$

$$ P \bigg(-z_{(0.005)} \leq \frac{\bar{X}_{1600}-18}{\frac{4}{3(1600)}} \leq z_{(0.005)} \bigg) = 0.99 $$

$$ P \bigg(18-\frac{4z_{(0.005)}}{3(1600)} \leq \bar{X}_{1600} \leq 18+\frac{4z_{(0.005)}}{3(1600)}\bigg) = 0.99 $$

We see that $ a = 18-\frac{4z_{(0.005)}}{3(1600)} $ and $ b = 18+\frac{4z_{(0.005)}}{3(1600)} $ are the lower and upper bounds, respectively.