Question

In: Statistics and Probability

1. Find the mean, standard deviation, and five number summaries in each type of M&M (plain...

1. Find the mean, standard deviation, and five number summaries in each type of M&M (plain and peanut) for the total number of M&Ms in each bag.
2. Find the mean, standard deviation, and five number summaries for the weights of Skittles in each bag.
3. Find the mean, standard deviation, and five number summaries for the weights of Plain M&Ms in each bag.
4. Use the information found in Step 1 to draw side by side boxplots, comparing the total number of M&Ms in each bag of plain and peanut M&Ms. Discuss the shape of each.
5. Use the information found in Steps 2 & 3 to draw side by side boxplots, comparing the weights of M&Ms in each bag of Plain and the weights in each bag of Skittles. Discuss the shape of each.
6. Write a summary paragraph about your data. Make at least 4 different observations about your data and/or boxplots using your statistical knowledge.

Plain M&M: 15,14,15,15,14,16,15,14,15,16,16,14,15,14,15,15,14,17,17,15,15,17,14,16,14,16,17,13,16,16, 15, 17,6,14,16,15,13,15,17,15,17,14,15,14,16,14

Peanut Butter: 8,6,8,7,8,8,8,7,3,7,10,7,7,6,8,7,8,7,7,7,8,8,8,7,9,7,8,8,7,8,8,8,6,7,8,8,8,7,6,7,7,8,8,8,14,8,8,7,6

Skittles: 15,14,15,14,14,15,15,14,15,15,14,15,15,15,14,15,20,15,15,15,14,14,15,15,15,16,12,15,15,14, 14,15,14,15,14,14,14,13,16

Solutions

Expert Solution

1)

Plain M&M:

mean and standard deviation

data data-mean (data - mean)2
15 0.0435 0.00189225
14 -0.9565 0.91489225
15 0.0435 0.00189225
15 0.0435 0.00189225
14 -0.9565 0.91489225
16 1.0435 1.08889225
15 0.0435 0.00189225
14 -0.9565 0.91489225
15 0.0435 0.00189225
16 1.0435 1.08889225
16 1.0435 1.08889225
14 -0.9565 0.91489225
15 0.0435 0.00189225
14 -0.9565 0.91489225
15 0.0435 0.00189225
15 0.0435 0.00189225
14 -0.9565 0.91489225
17 2.0435 4.17589225
17 2.0435 4.17589225
15 0.0435 0.00189225
15 0.0435 0.00189225
17 2.0435 4.17589225
14 -0.9565 0.91489225
16 1.0435 1.08889225
14 -0.9565 0.91489225
16 1.0435 1.08889225
17 2.0435 4.17589225
13 -1.9565 3.82789225
16 1.0435 1.08889225
16 1.0435 1.08889225
15 0.0435 0.00189225
17 2.0435 4.17589225
6 -8.9565 80.21889225
14 -0.9565 0.91489225
16 1.0435 1.08889225
15 0.0435 0.00189225
13 -1.9565 3.82789225
15 0.0435 0.00189225
17 2.0435 4.17589225
15 0.0435 0.00189225
17 2.0435 4.17589225
14 -0.9565 0.91489225
15 0.0435 0.00189225
14 -0.9565 0.91489225
16 1.0435 1.08889225
14 -0.9565 0.91489225

five number summary

Position XXX (Asc. Order)
1 6
2 13
3 13
4 14
5 14
6 14
7 14
8 14
9 14
10 14
11 14
12 14
13 14
14 14
15 14
16 15
17 15
18 15
19 15
20 15
21 15
22 15
23 15
24 15
25 15
26 15
27 15
28 15
29 15
30 15
31 16
32 16
33 16
34 16
35 16
36 16
37 16
38 16
39 16
40 17
41 17
42 17
43 17
44 17
45 17
46 17

Peanut Butter:

mean and standard deviation

data data-mean (data - mean)2
8 0.4694 0.22033636
6 -1.5306 2.34273636
8 0.4694 0.22033636
7 -0.5306 0.28153636
8 0.4694 0.22033636
8 0.4694 0.22033636
8 0.4694 0.22033636
7 -0.5306 0.28153636
3 -4.5306 20.52633636
7 -0.5306 0.28153636
10 2.4694 6.09793636
7 -0.5306 0.28153636
7 -0.5306 0.28153636
6 -1.5306 2.34273636
8 0.4694 0.22033636
7 -0.5306 0.28153636
8 0.4694 0.22033636
7 -0.5306 0.28153636
7 -0.5306 0.28153636
7 -0.5306 0.28153636
8 0.4694 0.22033636
8 0.4694 0.22033636
8 0.4694 0.22033636
7 -0.5306 0.28153636
9 1.4694 2.15913636
7 -0.5306 0.28153636
8 0.4694 0.22033636
8 0.4694 0.22033636
7 -0.5306 0.28153636
8 0.4694 0.22033636
8 0.4694 0.22033636
8 0.4694 0.22033636
6 -1.5306 2.34273636
7 -0.5306 0.28153636
8 0.4694 0.22033636
8 0.4694 0.22033636
8 0.4694 0.22033636
7 -0.5306 0.28153636
6 -1.5306 2.34273636
7 -0.5306 0.28153636
7 -0.5306 0.28153636
8 0.4694 0.22033636
8 0.4694 0.22033636
8 0.4694 0.22033636
14 6.4694 41.85313636
8 0.4694 0.22033636
8 0.4694 0.22033636
7 -0.5306 0.28153636
6 -1.5306 2.34273636

five point summary

we are allowed to solve one qustion only


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