Question

In: Statistics and Probability

Bag Blue Orange Green Yellow Red Brown Total Number of Candies 1 9 13 14 10...

Bag Blue Orange Green Yellow Red Brown Total Number of Candies
1 9 13 14 10 7 7 60
2 13 10 6 9 9 8 55
3 13 12 4 10 9 6 54
4 16 13 8 6 6 8 57
5 10 10 12 5 15 4 56
6 9 18 3 6 12 12 60
7 11 13 6 15 8 6 59
8 12 18 5 9 6 5 55
9 12 10 8 15 10 1 56
10 12 14 8 12 7 9 62
11 11 10 9 11 8 8 57
12 13 11 11 7 10 5 57
13 10 15 7 10 8 8 58
14 13 14 6 8 8 5 54
15 10 14 5 6 13 10 58
16 15 14 5 8 5 10 57
17 11 16 9 8 5 9 58
18 14 9 8 7 9 6 53
19 13 7 8 8 9 8 53
20 10 10 7 10 11 11 59
21 8 7 7 13 14 10 59
22 8 11 9 10 5 11 54
23 7 8 9 8 8 13 53
24 10 11 6 7 12 10 56
25 9 13 7 8 9 8 54
26 10 6 13 9 9 8 55
27 10 12 9 8 10 7 56
28 11 12 7 12 10 8 60
29 9 14 7 9 8 6 53
30 10 10 7 10 11 11 59

We evaluated the number and distribution of M&M Minis® candies. A sample of 30 bags was collected and the data is reported in the attached Excel file. This study examined the color percentages and the total number of candies in each bag.

If a random sample of 50 bags is selected, find the probability that the mean number of candies in a bag is at least 56 candies. Assume the total number of candies is normally distributed. F) Assume the total number of candies is normally distributed, find a 95% confidence interval for the mean total number of candies.

Solutions

Expert Solution

Let X denotes the total number of candies in a bag.

Since the distribution of X is normal.

X ~ N ( mu, sigma2).

Xbar = Mean number of candies in bag.

The distribution of Xbar is approximately normal.

Xbar ~ N (mu, sigma2/n).

Since Xbar = sum(Xi) / n and S2 = sum(Xi- Xbar)^2 / n-1 is an unbiased estimators of mu and sigma2 respectively.

Total Number of Candies (X) (Xi- Xbar)^2
60 11.7875
55 2.4545
54 6.5879
57 0.1877
56 0.3211
60 11.7875
59 5.9209
55 2.4545
56 0.3211
62 29.5207
57 0.1877
57 0.1877
58 2.0543
54 6.5879
58 2.0543
57 0.1877
58 2.0543
53 12.7213
53 12.7213
59 5.9209
59 5.9209
54 6.5879
53 12.7213
56 0.3211
54 6.5879
55 2.4545
56 0.3211
60 11.7875
53 12.7213
59 5.9209
1697 181.3667

S = sqrt(6.2540) = 2.5008

n = number of bags randomly selected = 50.

P ( Mean number of candies in a bag at least 56 candies) = P ( Xbar >= 56)

since

= P ( Z >= -1.6023) = 0.9455

P ( Mean number of candies in a bag at least 56 candies) = 0.9455.

Alpha: level of significance = 0.05.

Zalpha/2 = Z0.025 = 1.96

95% confidence interval for mean total number of candies is

( Xbar - (S/sqrt(n) * Zalpha/2, Xbar + (S/sqrt(n) * Zalpha/2)

= ( 56.5667 - ( 2.5008/ sqrt(50) * 1.96 , 56.5667 - ( 2.5008/ sqrt(50) * 1.96)

= ( 55.8735, 57.2599)

orchestra answered 2 years ago
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