Statistics for Decision-Making

9. Compare the standard deviation for the heights of males and the standard deviation for the heights of females in the class. Compare the values and explain what can be concluded based on the numbers. The females standard deviation was 2.5 (rounded from 2.533) and the males standard deviation was 4.2 (rounded from 4.21). This shows me that the male height was ranged out more. 10. Using the empirical rule, 95% of female heights should be between what two values? Either show work or explain how your answer was calculated. The female mean was 67.3 and the standard deviation was 2.5. According to the empirical rule 95% of female heights would be between + or – two standard deviations. This states: 67.3 – 2(2.5) and 67.3 + 2(2.5). 95% of female heights would be between 62.3 inches and 72.3 inches. 11. Using the empirical rule, 68% of male heights should be between what two values? Either show work or explain how your answer was calculated. The male mean was 70 and the standard deviation was 4.2. According to the empirical rule 68% of female heights would be between + or – two standard deviations. This states: 70 – 2(4.2) and 70 + 2(4.2). 68% of male heights would be between 61.6 inches and 78.4 inches.

MATH | 221 | Statistics for Decision-Making | lab | 9. Compare the standard deviat

y = -31.737x + 0.4668   Points Received: 0 of 8   Comments: ( T C O 9 ) A r e s e a r c h e r f o u n d a s i g n i f i c a n t r e l a t   i o n s h i p b e t w e e n a s t u d e n t ’ s I Q , a , t h e g r a d e p o i n t a v e r a g e , b , a n d t h e s c o r e , y , o n t h e v e r b a l s e c t i o n o f t h e S A T t e s t . T h e r e l a t i o n s h i p c a n b e r e p r e s e n t e d b y t h e m u l t i p l e r e g r e s s i o n e q u a t i o n y = 2 5 0 + 1 . 5 a + 8 0 b . P r e d i c t t h e S A T v e r b a l s c o r e ( t o t h e n e a r e s t w h o l e n u m b e r ) o f a s t u d e n t w h o s e I Q i s 1 3 0 a n d g r a d e p o i n t a v e r a g e i s 2 . 2 . Student Answer:  625   621 (Ch 9: y = 250 + (1.5 * 130) + (80 * 2.2))   421   371   Points Received: 0 of 8   Comments: ( T C O 9 ) I n t e r p r e t a n r v a l u e o f 0 . 9 2 .   Student Answer:  strong negative correlation   weak negative correlation   strong positive correlation (Chapter 9)   no correlation   Points Received: 3 of 3   Comments: Page: 1   2   Page: 1   2   1. Id en tif y th e cl as s wi dt h.   Student Answer:     Instructor Explanation: Chapter 2: (2 pts) The class width is 4; (2 pts) the midpoint of the ﬁrst class is 9.5; (2 pts) the class boundaries are 7.5 - 11.5; (4 pts) the relative frequencies are (18/158 = 0.1139), (23/158 = 0.1456), (38/158 = 0.2405), (47/158 = 0.2975), (32/158 = 0.2025)   Points Received: 0 of 10   Comments: ( T C O 3 ) T h e n u m b e r s o f h o m e r u n s t h a t S a m   Student Answer:  1. Stem/Leaf 0/4 1/0,5 2/5 3/3,6 4/0,9 5/0 6/3,4,6 2. 35.61 3. 36 4. 4,8,10,15,25,33,36,40,49,50,63,64,66 = no mode (there is not a number that repeats) 5. 62 6. 487.25 7. 22.07  Instructor Explanation: Chapter 2: (5 pts) The stem-and-leaf should look like this: 0 | 4 8 1 | 5 0 2 | 5 3 | 3 6 4 | 9 0 5 | 0 6 | 6 3 4 (1 pts) The mean is 35.6. (3 pts) The median is 36 (1 pts) There is no mode. (1 pts) The range is 62 (3 pts) The variance is 449.775 (using population formula since it is every single year of the ﬁrst 13 years) (1 pts) The standard deviation is 21.2   Points Received: 7 of 10   Comments: Page: 1   2  

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