91Ó°ÊÓ

It is given that the mean velocity is 936 feet per second, and the standard deviation is 10 feet per second.

i.e.,$\stackrel{¯}{\mathrm{x}}=936\mathrm{and}\mathrm{s}=10$

Here, Chebyshev’s rule can be used to describe the distribution.

Since Chebyshev’s rule states that at least $\frac{3}{4}$of the data will fall within 2 standard deviations of the mean.

Therefore,

$$\begin{gathered} \stackrel{¯}{x}+2s=936Â\pm \left(2\right)\left(10\right) \\ =936Â\pm 20 \\ =\left(936-20,936+20\right) \\ =916,956 \end{gathered}$$

Thus, at least $\frac{3}{4}$of the data will fall in the interval (916,956).

Also, Chebyshev’s rule states that at least $\frac{8}{9}$of the data will fall within 3 standard deviations of the mean.

Therefore,

$$\begin{gathered} \stackrel{¯}{x}+3s=936Â\pm \left(3\right)\left(10\right) \\ =936Â\pm 30 \\ =\left(936-30,936+30\right) \\ =\left(906,966\right) \end{gathered}$$

Thus, at least $\frac{8}{9}$of the data will fall in the interval (906,966).

If an unknown bullet’s velocity is measured as 1,000 feet per second, it would be improbable that a well-known company manufactures this bullet. Its velocity would be almost 6 standard deviations above the mean of 936 if it were from our manufacturer, which is an improbable event.