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

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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lamusesamuset

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