Question

If a population data set is normally distributed, what isthe proportion of measurements you would expect to fallwithin the following intervals?

$$\begin{gathered} \mathbf{a}\mathbf{.}\mathbf{}\mathbf{\mu }\mathbf{\pm }\mathbf{\sigma } \\ \mathbf{b}\mathbf{.}\mathbf{}\mathbf{\mu }\mathbf{\pm }\mathbf{2}\mathbf{\sigma } \\ \mathbf{c}\mathbf{.}\mathbf{}\mathbf{\mu }\mathbf{\pm }\mathbf{3}\mathbf{\sigma } \end{gathered}$$

Short answer

a. The proportion of measurements that fall in the interval is 68%

b.The proportion of measurements that fall in the interval is 95%

c.The proportion of measurements that fall in the interval is 100%

Step-by-step solution

Given Information

The population is normally distributed.

Calculating the proportion when μ±2σ

b.

From the property of normal distribution,

$$\begin{gathered} P\left(\mu -\sigma <X<\mu +\sigma \right)=P\left(\frac{\mu -\sigma -\mu }{\sigma }<\frac{X-\mu }{\sigma }<\frac{\mu -\sigma -\mu }{\sigma }\right) \\ =P\left(-1<Z<1\right) \\ =P\left(Z<1\right)-P\left(Z\le -1\right) \\ =0.84134-0.15866 \\ =0.68269 \\ 𝆏0.68 \\ P\left(\mu -\sigma <X<\mu +\sigma \right)𝆏0.68 \end{gathered}$$

The proportion of measurements that fall within 1 standard deviation of the mean is approximately 68%.

Calculating the proportion when

$$\begin{gathered} P\left(\mu -2\sigma <X<\mu +2\sigma \right)=P\left(\frac{\mu -2\sigma -\mu }{\sigma }<\frac{X-\mu }{\sigma }<\frac{\mu -2\sigma -\mu }{\sigma }\right) \\ =P\left(-2<Z<2\right) \\ =P\left(Z<2\right)-P\left(Z\le -2\right) \\ =0.97725-0.02275 \\ =0.9545 \\ 𝆏0.95 \\ P\left(\mu -2\sigma <X<\mu +2\sigma \right)𝆏0.95 \end{gathered}$$

The proportion of measurements that fall within 2 standard deviations of the mean is approximately95%.

Calculating the proportion when

c.

$$\begin{gathered} P\left(\mu -3\sigma <X<\mu +3\sigma \right)=P\left(\frac{\mu -3\sigma -\mu }{\sigma }<\frac{X-\mu }{\sigma }<\frac{\mu -3\sigma -\mu }{\sigma }\right) \\ =P\left(-3<Z<3\right) \\ =P\left(Z<3\right)-P\left(Z\le -3\right) \\ =0.99865-0.00135 \\ =0.9973 \\ 𝆏1 \\ P\left(\mu -3\sigma <X<\mu +3\sigma \right)𝆏1 \end{gathered}$$

The proportion of measurements that fall within 3 standard deviations of the mean is approximately 100%.