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Sagot :
Given that X follows a normal distribution with a mean of 75 and a standard deviation of 5.
To Determine: P(X>80)
Solution:
P(X>a)
[tex]P(X>a)=P(\frac{X-\mu}{\sigma}>\frac{a-\mu}{\sigma})[/tex]P(X>80)
[tex]\begin{gathered} P(X>80)=P(\frac{X-\mu}{\sigma}>\frac{80-\mu}{\sigma}) \\ \mu=75,\sigma=5 \end{gathered}[/tex][tex]P(X>80)=P(\frac{X-\mu}{\sigma}>\frac{80-75}{5})[/tex][tex]\begin{gathered} P(X>80)=P(Z>\frac{5}{5}) \\ P(X>80)=P(Z>1) \end{gathered}[/tex]We now go to the standard normal distribution table to look up P(Z>1) and for Z=1.00
We find that
[tex]P(Z<1)=0.8413[/tex]Note, however, that the table always gives the probability that Z is less than the specified value, i.e., it gives us P(Z<1)=0.8413
Therefore:
[tex]\begin{gathered} P(Z>1)=1-P(Z<1) \\ P(Z>1)=1-0.8413 \\ P(Z>1)=0.1587 \end{gathered}[/tex]In percentage
[tex]\begin{gathered} P(X>80)=1-P(Z<1)=0.1587 \\ In\text{ Percentage} \\ P(X>80)=0.1587\times\frac{100}{1}\text{ \%=15.87\%} \end{gathered}[/tex]Hence, P(X>80) = 15.87%
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