Question

12. Below is a sample of weights (in kilograms) of incoming freshmen in a college: 729774 93 68 5 64 56 76 s8 s0 p 61 58 70 s1 58 92 57 57 a) Construct a stem-and-leaf plot to assess whether the data are from an approximately normal distribution. (2 pts) 100 12. У Compute the standard deviation for the sample data (2 pts) b) Use the values of Q, Qu and s to assess whether the data come from an approximately normal distribution. (3 pts) c) d) Use Empirical Rule to assess whether the data are from an approximately normal distribution. (Recall that according to the rule, for a data set to come from an approximately normal distribution, " 68% of data are 1 standard deviation about the mean, 95% of data are 2 standard deviations about the mean, and 100% of data are 3 standard deviations about the mean) (3 pts)

Answer 1

a) The stem and leaf plot is given below :

5 | 0366789
6 | 14778
7 | 00124
8 |
9 | 237

The decimal point is 1 digit(s) to the right of the |

b)

	x	x - $\dpi{80} \fn_jvn \bar{x}$	( x - $\dpi{80} \fn_jvn \bar{x}$ )2
	72	4.25	18.0625
	97	29.25	855.5625
	74	6.25	39.0625
	93	25.25	637.5625
	68	0.25	0.0625
	59	-8.75	76.5625
	64	-3.75	14.0625
	56	-11.75	138.0625
	70	2.25	5.0625
	58	-9.75	95.0625
	50	-17.75	315.0625
	71	3.25	10.5625
	67	-0.75	0.5625
	56	-11.75	138.0625
	70	2.25	5.0625
	61	-6.75	45.5625
	53	-14.75	217.5625
	92	24.25	588.0625
	57	-10.75	115.5625
	67	-0.75	0.5625
Total	271		3315.75

The mean $\dpi{80} \fn_jvn \bar{x} = \frac{271}{4} = 67.75$

The standard deviation is given by the formula :

$\dpi{80} \fn_jvn s =\sqrt{\frac{\sum (x - \bar{x})^{2}}{n-1}}$

$\dpi{80} \fn_jvn s =\sqrt{\frac{3315.75}{20-1}} = 13.2103 \approx 13$

d)  The empirical rule states that for a normal distribution :

• 68% of data falls within the first standard deviation from the mean.
• 95% fall within two standard deviations.
• 99.7% fall within three standard deviations.

In this context we have $\dpi{80} \fn_jvn \bar{x} = 67.75$ and $\dpi{80} \fn_jvn s= 13.2103$

Applying empirical rule to given data we get:

• 68% of data falls within 67.75 $\dpi{80} \fn_jvn \pm$ 13.2103 i.e. (54.5397, 80,9603 )
• 95% fall within 67.75 $\dpi{80} \fn_jvn \pm$ 2*13.2103 i.e. (42.3294 , 95.1706 )
• 99.7% fall within 67.75 $\dpi{80} \fn_jvn \pm$ 3*13.2103 i.e. (29.1191 , 108.3809 )

From the data 16 observation are within (54.5397, 80,9603 ) i.e. about (16/20)*100 = 80% observation are within (54.5397, 80,9603 ).

And 19 observation are within (42.3294 , 95.1706 ) i.e. about (19/20)*100 = 95% observation are within (42.3294 , 95.1706 ).

And 20 observation are within (29.1191 , 108.3809 ) i.e. about (100/20)*100 = 100% observation are within (29.1191 , 108.3809 )

Hence empirical rule is satisfied here.