Chapter 10.3, Problem 26PFA

To determine

To select the correct statements.

Answer to Problem 26PFA

A and B are correct.

Explanation of Solution

Given information: Data set is

5.87, 5.47, 5.02, 4.62, 6.83, 6.2, 6.15, 5.55, 3.05, 5.09, 5.07

Formula used: Standard deviation formula is,

  ${\sigma }_{}=\sqrt{\frac{{\displaystyle \sum _{}^{}{(x_{}^{}-\mu )}^2}}{N}}$

Lower quartile is,

  $=\frac{(N+1)}{4}$

Upper quartile is,

  $=\frac{3(N+1)}{4}$

Inter-quartile range is,

  $IQR=Q3-Q1$

Median is,

  $=\frac{(N+1)}{2}$

Calculation:

Sorting data,

3.05,4.62,5.02,5.07,5.09,5.47,5.55,5.87,6.15,6.2,6.83

Verifying the options,

For option A,

3.05 is a value very far from the other data points. It is an outlier

Hence, A is correct.

For option B,

For Mean substituting values,

= $\frac{3.05+4.62+5.02+5.07+5.09+5.47+5.55+5.87+6.15+6.2+6.83}{11}$

Solving,

=5.35

Value	$x_{}^{}-\mu$	${(x_{}^{}-\mu )}^2$
3.05	-2.306363636	5.319313221
4.62	-0.7363636364	0.542231405
5.02	-0.3363636364	0.1131404959
5.07	-0.2863636364	0.08200413225
5.09	-0.2663636364	0.0709495868
5.47	0.1136363636	0.01291322313
5.55	0.1936363636	0.03749504131
5.87	0.5136363636	0.263822314
6.15	0.7936363636	0.6298586776
6.2	0.8436363636	0.711722314
6.83	1.473636364	2.171604133
Total	${\displaystyle \sum _{}^{}{(x_{}^{}-\mu )}^2}$ =9.995

For Standard deviation substituting the value,

  $=\frac{9.995}{11}$

On solving

=0.95

B is correct.

For option C.

For lower quartile substituting the value,

  $=\frac{(11+1)}{4}$

So,

  $=\frac{12}{4}$

On solving,

  $=3^{rd}ob$

On substituting the value,

  $Q1=5.02$

For upper quartile substituting the value,

  $=\frac{3(11+1)}{4}$

So,

  $=\frac{3\times 12}{4}$

On solving,

  $=9^{th}ob$

On substituting the value,

  $Q3=6.15$

For Interquartile range. Substituting the value,

  $=6.15-5.02$

On solving,

  $=1.13$

Hence, C is incorrect.

For option D

For Median substituting the value,

  $=\frac{11+1}{2}$

On solving,

  $=6^{th}ob$

On substituting the value,

  $median=5.47$

Hence, D is incorrect.

For option E

The measure of central tendency most affected by an outlier is mean.

Hence, E is incorrect