Check the solution to "Let X1,X2,...,Xn be n independent random variables each..."

vazelinahS

Answered question

2021-06-04

Let

X_{1}, X_{2}, . . ., X_{n}

be n independent random variables each with mean 100 and standard deviation 30. Let X be the sum of these random variables.
Find n such that

P r (X > 2000) \geq 0.95

Answer & Explanation

unessodopunsep

Skilled2021-06-05Added 105 answers

Step 1
Given that $μ = 100$ and $σ = 30$
$E (X) = E (\sum X_{i}) = n μ = 100 n$
$σ (X) = σ (\sum X_{i}) = \sqrt[30]{n}$
Step 2
Consider,
$P (X > 2000) \geq 0.95$
$P (\frac{X - E (X)}{σ (X)} > \frac{2000 - 100 n}{\sqrt[30]{n}}) \geq 0.95$
$P (z > \frac{200 - 10 n}{\sqrt[3]{n}}) \geq 0.95$
$1 - P (z \leq \frac{200 - 10 n}{\sqrt[3]{n}}) \geq 0.95$
$1 - 0.95 \geq P (z \leq \frac{200 - 10 n}{\sqrt[3]{n}})$
$P (z \leq \frac{200 - 10 n}{\sqrt[3]{n}}) \leq 0.05$
$\frac{200 - 10 n}{\sqrt[3]{n}} \leq - 1.645$
$200 - 10 n \leq 4.935 \sqrt{n}$
$4.935 \sqrt{n} - 10 n + 200 \leq 0$
$\sqrt{n} \leq - 4.23$
$n = 17.89$
$\approx 18$
Thus, $n = 18.$

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