SOLUTION: A survey of 52 U.S. supermarkets yielded the following relative frequency table, where X is the number of checkout lanes at a randomly chosen supermarket. x 1 2 3 4 5 6 7 8 9 10

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Question 1196769: A survey of 52 U.S. supermarkets yielded the following relative frequency table, where X is the number of checkout lanes at a randomly chosen supermarket.
x 1 2 3 4 5 6 7 8 9 10
P(X = x)
0.01 0.02 0.02 0.10 0.10 0.15 0.25 0.20 0.10 0.05
(a)
Compute 𝜇 = E(X). HINT [See Example 3.]
E(X) =
Interpret the result.
This was the most frequently observed number of checkout lanes in surveyed supermarkets.
There were, on average, this many checkout lanes in a supermarket that was surveyed.
There were at least this many checkout lanes in each supermarket that was surveyed.
There were at most this many checkout lanes in each supermarket that was surveyed.
Correct: Your answer is correct.
(b)
Find
P(X < 𝜇) or P(X > 𝜇).
P(x < 𝜇)
=
P(x > 𝜇)
=
Interpret the result.
Then
P(x > 𝜇)
is
greater than

Correct: Your answer is correct.

P(x < 𝜇).
Thus, most supermarkets have
more than

Correct: Your answer is correct.
the average number of checkout lanes.

Answer by ewatrrr(24785)   (Show Source): You can put this solution on YOUR website!

Hi
Using an Excel Spreadsheet
# 	p	p*(#)	
1	0.01	0.01	
2	0.02	0.04	
3	0.02	0.06	
4	0.1	0.4	
5	0.1	0.5	
6	0.15	0.9	
7	0.25	1.75	
8	0.2	1.6	
9	0.1	0.9	
10	0.05	0.5	
	1	6.66 sum 
μ = E(Number of checkout Lanes) = 7 


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