SOLUTION: 1. The Marketing Manager at Smart Post Delivery Company is analyzing the performance of a new product, The Mega Box. This is a very sturdy padded box inside which any item can be p

Question 806821: 1. The Marketing Manager at Smart Post Delivery Company is analyzing the performance of a new product, The Mega Box. This is a very sturdy padded box inside which any item can be placed and shipped without breaking. The manager has collected data from past delivery reports and found that 1% of the items placed in the Mega Box broke during shipping.
(i) If a random sample of 12 items placed in the Mega Box was selected, what is the probability that at most two items would break during shipping?
(ii) What is the mean number of items that would break during shipping?
2. Smart Snacks recently introduced a new snack product to its customers, the Chocolate Crisp Sandwich. The number of sandwiches sold over the last 5 days is described by the following probability distribution.
Number of sandwiches, x 15 20 25 30 35
Probability, P(x) 0.25 0.40 0.15 0.12 *
.
(i) What is the probability that 35 sandwiches were sold.
(ii) What is the probability that less than 25 sandwiches were sold.
(iii) Calculate the expected number of sandwiches to be sold.
(iv) Calculate the standard deviation for the number of sandwiches sold.

Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
The manager has collected data from past delivery reports and found that 1% of the items placed in the Mega Box broke during shipping.
(i) If a random sample of 12 items placed in the Mega Box was selected, what is the probability that at most two items would break during shipping?
Binomial Problem with n = 12 and p(broke) = 0.01
P(0<= x <=2) = binomcdf(12,0.01,2) = 0.9999
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(ii) What is the mean number of items that would break during shipping?
mean = np = 12*0.01 = 0.12
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2. Smart Snacks recently introduced a new snack product to its customers, the Chocolate Crisp Sandwich. The number of sandwiches sold over the last 5 days is described by the following probability distribution.
Number of sandwiches, x 15 20 25 30 35
Probability, P(x) 0.25 0.40 0.15 0.12 *
.
(i) What is the probability that 35 sandwiches were sold.
Ans:: 1 -(0.25+0.40+0.15+0.12) = 0.08
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(ii) What is the probability that less than 25 sandwiches were sold.
Ans: 0.25+0.40 = 0.65
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(iii) Calculate the expected number of sandwiches to be sold.
Ans: E(x) = 15*0.25+20*0.40+25*0.15+30*0.12+35*0.08 = 21.9
----------------
(iv) Calculate the standard deviation for the number of sandwiches sold.
Ans:: std = 7.9057
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Cheers,
Stan H.

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