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Find P (X > 80) if X follows a normal distribution with a mean of 75 and a standard deviation of 5.15.87%54.29%70.46%84.13%

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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)


P(X>a)=P((X-\mu)/(\sigma)>(a-\mu)/(\sigma))

P(X>80)


\begin{gathered} P(X>80)=P((X-\mu)/(\sigma)>(80-\mu)/(\sigma)) \\ \mu=75,\sigma=5 \end{gathered}
P(X>80)=P((X-\mu)/(\sigma)>(80-75)/(5))
\begin{gathered} P(X>80)=P(Z>(5)/(5)) \\ P(X>80)=P(Z>1) \end{gathered}

We now go to the standard normal distribution table to look up P(Z>1) and for Z=1.00

We find that


P(Z<1)=0.8413

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:


\begin{gathered} P(Z>1)=1-P(Z<1) \\ P(Z>1)=1-0.8413 \\ P(Z>1)=0.1587 \end{gathered}

In percentage


\begin{gathered} P(X>80)=1-P(Z<1)=0.1587 \\ In\text{ Percentage} \\ P(X>80)=0.1587*(100)/(1)\text{ \%=15.87\%} \end{gathered}

Hence, P(X>80) = 15.87%

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