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If the hight of $300$ students are normally distributed with mean $64.5$ inches and standard deviation $3.3$ inches,find the hight below which $99\%$ of the student lie.

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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
  • $Z=\large\frac{X-\mu}{\sigma}$
Step 1:
Let $X$ be the random variable denoting the height of a student .
$X\sim N(64.5,(3.3)^2)$
Step 2:
To find the height below which 99% of the students lie.
Let $X_1$ be the height.
$P(X < X_1)=0.99$
Step 3:
Let $Z$ be the standard normal variate.
$Z=\large\frac{X-\mu}{\sigma}$
$\;\;\;=\large\frac{X-64.5}{3.3}$
Step 4:
When $X=X_1$
$Z=\large\frac{X_1-64.5}{3.3}$
$P(Z < Z_1)=0.99$
$\Rightarrow P(Z< Z_1)=0.5+0.49$
$\Rightarrow P(0< Z < Z_1)=0.49$
$Z=2.33$(from the table)
Step 5:
$Z=\large\frac{X_1-\mu}{\sigma}$
$\mu =64.5$
$2.33=\large\frac{X_1-64.5}{3.3}$
$X_1-64.5=(2.33)(3.3)$
$X_1=64.5+(2.33)(3.3)$
$\;\;\;\;=64.5+7.7$
$\;\;\;\;=72.2$inches
answered Sep 19, 2013 by sreemathi.v
 

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