Question

	Exercise 52:		A consumer watchdog organization is concerned about credit card debt. A survey of 10 young
			adults with credit card debt of more than $2,000 showed they paid an average of just over
			$100 per month against their balances. Listed below is the amounts each young adult paid
			last month.			110	116	103	93	99
						113	87	101	109	104
			Determine the sample standard deviation.						[answer is the same using STDEV or STDEV.S]
	Exercise 54:		The mean of a group of sample observations is 500 & the standard deviation is 50.
			According to Chebyshev's theorem, at least what percentage of the data will lie between 400 & 600?

Answer 1

52)The sample mean is

$\dpi{100} \fn_jvn \overline{X}=\frac{X_1+X_2+...+X_{10}}{10}=103.5\\$

The sample standard deviation is

$\dpi{100} \fn_jvn s^2=\frac{\left (X_1 -\overline{X} \right )+\left (X_2 -\overline{X} \right )^2+...+\left (X_{10} -\overline{X} \right )^2}{10-1}=8.9969\\$

54)The Chebyshev's inequality is

$\dpi{100} \fn_jvn P(\left | X-\mu \right |\geqslant k)\leqslant \frac{\sigma ^2}{k^2}$

Which is equivalent to,

$\dpi{100} \fn_jvn -P(\left | X-\mu \right |\geqslant k)\geqslant -\frac{\sigma ^2}{k^2}\\ 1-P(\left | X-\mu \right |\geqslant k)\geqslant 1- \frac{\sigma ^2}{k^2}\\ P(\left | X-\mu \right |< k)\geqslant 1- \frac{\sigma ^2}{k^2}$

We need to find

$\dpi{100} \fn_jvn P(400<X<600)=P(400-500<X-500<600-500)\\ P(400<X<600)=P(-100<X-500<100)\\ P(400<X<600)=P(-100<X-\mu <100)\\ P(400<X<600)=P(\left |X-\mu \right |<100)$

Now applying Chebyshev bound,

$\dpi{100} \fn_jvn P(400<X<600)=P(\left |X-\mu \right |<100)\\ P(400<X<600)\geqslant 1- \frac{50 ^2}{100^2}\\ P(400<X<600)\geqslant \frac{3}{4}$

So at least $\dpi{100} \fn_jvn 75\%$ of data will lie between 400 and 600.