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There're 3 brothers who are saving money every month. The 1st brother saves \$160 monthly; the savings of the 2nd brother obey normal distribution with $E=\$280$ ($E$ = expected value), $\sigma = 30$ ($\sigma$ = standard deviation); the savings of the 3rd brother are 1.5 times bigger than that of the 2nd brother. The monthly savings are independent in different months.

The parents of the brothers decide to add \$100 each month when the 1st and the 2nd brother save more than the 3rd brother. What is the probability that in a given month the parents will add money?

Let "brother" = $B$, then I think I need to calculate $P(B_3

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    third brother is bigger than the second brother probably, not himself – 2017-02-02
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    @Momo all I know is that $E(B_3)>E(B_2)$ but we need to find the probability of the event $P(B_3 – 2017-02-02
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    I'm talking about "the savings of the 3rd brother are 1.5 times bigger than that of the 3nd brother" – 2017-02-02
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    Momo has pointed out a probable typo in the last line of the first paragraph, conjecturing, if my understanding is correct, that '3nd' should be '2nd'. – 2017-02-02
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    that's right but when I calculate the probability how do we know how much borthers saved in a given month? should I just assume they mean expected values? – 2017-02-02
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    There is only one random thing here, which is $B_2$. I already solved it for you in the answers. – 2017-02-02
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    @Jan true it was a typo – 2017-02-02

2 Answers 2

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$B_1$ is a constant (equals 160 with probability 1)

Now:

$$P(B_1+B_2>B_3)=P(160+B_2>1.5 B_2)=P(0.5 B_2<160)=P(B_2<320)$$

and i think you can solve the rest yourself.

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    how can we assume that 3rd brother's savings also obey normal distribution ? They just said that the expected value is 1.5 times bigger than that of the second brother – 2017-02-02
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    and also should I just assume in these types of questions that they assume expected values of variables? because each brother can save unpredictable and different amounts of money each month – 2017-02-02
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    I haven't really used that the third brother's savings obey a normal distribution in the solution, but the answer is yes. In other words if $X\sim N(\mu,\sigma^2)$ and $a$ is a constant then $aX\sim N(a\mu,a^2\sigma^2)$. And no, the problem is not about expectations. The problem is about the probability of a normal random variable to take values in a given interval, in your case $(-\infty, 320)$ – 2017-02-02
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    so what are the values you plug inside $P()$? I see them as expected values but I don't understand why we need to base our calculation on expected values for this problem because one month 3rd brother can save more money than the other 2 theoretically – 2017-02-02
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    They are random variables, not expected values. It is like rolling a die. Do you get a random variable or an expected value? Think about a similar problem like: you roll a die each month and if face "6" comes up, your parents give to you $100. – 2017-02-02
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From the question, $B_{1}=160$, $B_{2}\sim\mathcal{N}(280,30^2)$ and $B_{3}=1.5B_{2}$. You are looking for $\mathbb{P}[B_{1}+B_{2}\geq B_{3}]=\mathbb{P}[160+B_{2}\geq 1.5B_{2}]=\mathbb{P}[B_{2}\leq320]$. Since $B_{2}\sim\mathcal{N}(280,30^{2})$, $\frac{B_{2}-280}{30}\sim\mathcal{N}(0,1)$. Hence $\mathcal{P}[B_{2}\leq320]=\mathcal{P}[\frac{B_{2}-280}{30}\leq\frac{4}{3}]=\Phi(\frac{4}{3})\approx0.91$, where $\Phi$ is cdf of standard normal distribution.