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If X is normal with mean 12 and standard deviation 2 , find P(10<X15)P(10<X \leq 15).


Answered 2 years ago
Answered 2 years ago
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We are given the normal random variable XX whose mean has value μ=12\mu=12 and standard deviation has value σ=2\sigma=2. Goal is to calculate the probability P(10<X15)P(10<X\leq15). Let's try to standardize this probability. For any normal random variable with mean μ\mu and standard deviation σ\sigma, the following equality is true:

P(c1<X<c2)=P(c1μσ<Xμσ<c2μσ)=P(c1μσ<Z<c2μσ)\begin{aligned} % % Remove numbering (before each equation) P(c_1<X<c_2) &=& P\left(\frac{c_1-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{c_2-\mu}{\sigma}\right) \\ &=& P\left(\frac{c_1-\mu}{\sigma}<Z<\frac{c_2-\mu}{\sigma}\right) \end{aligned}

Since we have values for mean and standard deviation, we get that:

P(10<X15)=P(10122<X0.050.0115122)=P(22<Z32)=P(1<Z1.5)\begin{aligned} % % Remove numbering (before each equation) \textcolor{#c34632}{P(10<X\leq15)} &=& P\left(\frac{10-12}{2}<\frac{X-0.05}{0.01}\leq\frac{15-12}{2}\right) \\ &=& P\left(\frac{-2}{2}<Z\leq\frac{3}{2}\right) \\ &=& \textcolor{#c34632}{P\left(-1<Z\leq1.5\right)} \end{aligned}

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