A normal distribution has a mean of 50 and a standard deviation of 4. Determine

Jaden Easton 2021-10-26 Answered
A normal distribution has a mean of 50 and a standard deviation of 4. Determine the value below which 95 percent of the observations will occur.

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Expert Answer

lamusesamuset
Answered 2021-10-27 Author has 17816 answers

Mean of the normal distribution is \(\displaystyle\mu={50}\) and the standard deviation is \(\displaystyle\sigma={4}\).
Let X be the random variable.
Now,
\(\displaystyle{z}={\frac{{{X}-\mu}}{{\sigma}}}\)
\(\displaystyle\Rightarrow{z}={\frac{{{X}-{50}}}{{{4}}}}\)
and \(\displaystyle{z}_{{{1}}}={\frac{{{X}_{{{1}}}-{50}}}{{{4}}}}\)
95 percent of the observations lie below \(\displaystyle{X}_{{{1}}}\).
\(\displaystyle\Rightarrow{P}{\left({z}{<}{z}_{{{1}}}\right)}={0.95}\)
\(\displaystyle\Rightarrow{0.5}+{P}{\left({0}{<}{z}{<}{z}_{{{1}}}\right)}={0.95}\)
\(\displaystyle\Rightarrow{P}{\left({0}{<}{z}{<}{z}_{{{1}}}\right)}={0.95}-{0.5}\)
\(\displaystyle\Rightarrow{P}{\left({0}{<}{z}{<}{z}_{{{1}}}\right)}={0.45}\)
\(\displaystyle\Rightarrow{z}_{{{1}}}={1.64}\) (from Appendix B1.)
Now,
\(\displaystyle{z}_{{{1}}}={\frac{{{X}_{{{1}}}-{50}}}{{{4}}}}\)
\(\displaystyle\Rightarrow{1.64}={\frac{{{X}_{{{1}}}-{50}}}{{{4}}}}\)
\(\displaystyle\Rightarrow{X}_{{{1}}}-{50}={6.56}\)
\(\displaystyle\Rightarrow{X}_{{{1}}}={50}+{6.65}\)
\(\displaystyle\Rightarrow{X}_{{{1}}}={56.65}\)
95 percent of the observations will lie below 56.65.
Result:
56.65

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