Advanced Normal Distribution Problem
Suppose that for a certain random variable\[X\]
normally distributed,\[P( 20 \le X \le 30)=0.3\]
\[P(X \le 30)=3 P(X \le 20)\]
We can write the first equation as
\[P(X \le 30)-P(X \le 20)=0.3\]
The simultaneous equations can be written
jatex options:inline}P(X \le 30)-P(X \le 20)=0.3{/jatex} (1)
\[P(X \le 30)-3 P(X \le 20)=0\]
(2)(1)-(2) gives
\[P(X \le 20)=0.3 \rightarrow P(X \le 20)=0.15\]
then \[P(X \le 30)=3 P(X \le 20)=3 \times 0.15 = 0.45 \]
.Now we have the equations
\[P(X \le 20)=0.15, \; P(X \le 20)= 0.45 \]
.Using normal distribution tables or a calculator gives
\[\frac{20- \mu}{\sigma}=-1.036, \; \frac{30 - \mu}{\sigma}=-0.125\]
Multiplying by
\[\sigma\]
and subtracting gives \[-10=-0.911 \sigma \rightarrow \sigma = \frac{10}{0.911}=10.97\]
then \[\mu=31.37\]
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