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Steele Electronics Inc. sells expensive brands of stereo equipment in several shopping malls.
The marketing research department of Steele reports that 37% of the customers entering the store that indicate they are browsing will,
in the end, make a purchase. Let the last 26 customers who enter the store be a sample.
a. How many of these customers would you expect to make a purchase? (Round the final answer to the nearest whole number.)
b. What is the probability that exactly six of these customers make a purchase? (Round the final answer to 4 decimal places.)
c. What is the probability 15 or more make a purchase? (Round the final answer to 4 decimal places.)
d. Does it seem likely at least one will make a purchase ("likely" refers if the probability is more than 70%)?
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(a) Based on given information, I expect that 37% of 26 customers, i.e. 0.37*26 = 10 (9.62 rounded) will make a purchase. ANSWER
For parts b), c) and d) notice that they are binomial distribution problems
with the number of trial 26, the probability of success of 37% = 0.37.
(b) In this case, I need calculate P(n=26; k = 6; p=0.37).
To facilitate calculations, I use an online (free of charge) calculator at this web-site
https://stattrek.com/online-calculator/binomial.aspx
It provides nice instructions and a convenient input and output for all relevant options/cases.
P(n=26; k = 6; p=0.37) = 0.05730377685, or 0.0573 (rounded). ANSWER
(c) In this case, I need calculate P(n=26; k >= 15; p=0.37).
I use the same online calculator. P(n=26; k >= 15; p=0.37) = 0.9573613397 or 0.9574 (rounded). ANSWER
(d) In this case, I need calculate P(n=26; k >= 1; p=0.37).
I use the same online calculator. P(n=26; k >= 1; p=0.37) = 0.99999393467 or 0.9999 (rounded). ANSWER
The problem is just solved: all questions are answered.