excans:Bb02fc7847: Difference between revisions
From Stochiki
(Created page with "'''Solution: B''' Let <math>X_1, \ldots, X_n </math> denote the life spans of the <math>n</math> light bulbs purchased. Since these random variables are independent and norma...") |
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0.9772 = | 0.9772 = \operatorname{P}[S > 40] = \operatorname{P}[ \frac{S-3n}{\sqrt{n}} > \frac{40-3n}{\sqrt{n}}]. | ||
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-2 \geq \frac{40-3n}{\sqrt{n}} | -2 \geq \frac{40-3n}{\sqrt{n}}. | ||
</math> | </math> | ||
Latest revision as of 16:21, 9 May 2023
Solution: B
Let [math]X_1, \ldots, X_n [/math] denote the life spans of the [math]n[/math] light bulbs purchased. Since these random variables are independent and normally distributed with mean 3 and variance 1, the random variable [math]S = X_1 + \cdots + X_n [/math] is also normally distributed with mean [math]\mu = 3n [/math] and standard deviation [math] \sigma = \sqrt{n}[/math]. Now we want to choose the smallest value for [math]n[/math] such that
[[math]]
0.9772 = \operatorname{P}[S \gt 40] = \operatorname{P}[ \frac{S-3n}{\sqrt{n}} \gt \frac{40-3n}{\sqrt{n}}].
[[/math]]
This implies that [math]n[/math] should satisfy the following inequality:
[[math]]
-2 \geq \frac{40-3n}{\sqrt{n}}.
[[/math]]
To find such an [math]n[/math], let’s solve the corresponding equation for [math]n[/math]:
[[math]]
\begin{align*}
-2 = \frac{40-3n}{\sqrt{n}} \\
-2\sqrt{n} = 40 -3n \\
3n-2\sqrt{n} -40 = 0 \\
(3\sqrt{n}+10)(\sqrt{n} -4) = 0 \\
\sqrt{n} = 4 \\
n = 16
\end{align*}
[[/math]]
Copyright 2023. The Society of Actuaries, Schaumburg, Illinois. Reproduced with permission.