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Suppose that the amount of cosmic radiation to which a person is exposed when flying by jet across the united states is a random variable having a normal distribution with a mean of $4.35 m$ rem and standard deviation of $0.59m$ rem. What is the probability that a person will be exposed to than $5.20m$ rem of cosmic radiation of such flight.

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Toolbox:
  • Standard normal distribution:
  • In a standard normal distribution $\mu=0,\sigma ^2=1$
  • The random variable $X$ can be converted to the standard normal variable $Z$ by the transformation
  • $Z=\large\frac{X-\mu}{\sigma}$
  • The probability density function $Z$ is $\phi(z)=\large\frac{1}{\sqrt{2\pi}}$$e^{-\Large\frac{1}{2}z^2};-\infty < Z < \infty$
  • $Z\sim N(0,1)$
Step 1:
Let $X$ denote the amount of cosmic radiation and a person is exposed to $X \sim N(4.35,(0.59)^2))$
To find $P(X > 5.20)$
Step 2:
Let $Z$ be the standard normal variate,
$Z=\large\frac{X-\mu}{\sigma}$
When $X=5.20$
$Z=\large\frac{5.20-4.35}{0.59}$
$\;\;\;=\large\frac{0.85}{0.59}$
$\;\;\;=1.44$
Step 3:
$\therefore P(X > 5.20)=P(Z > 1.44)$
$\qquad\qquad\qquad=P(0 < Z < \infty)-P(0 < Z < 1.44)$
$\qquad\qquad\qquad=0.5-0.4251$
$\qquad\qquad\qquad=0.0749$
answered Sep 19, 2013 by sreemathi.v
 

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