Question 1196767: 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.
Find
P(X < 𝜇) or P(X > 𝜇).
P(x < 𝜇)
=
P(x > 𝜇)
=
Interpret the result.
Then
P(x > 𝜇)
is
greater than
.
P(x < 𝜇).
Thus, most supermarkets have
more than
.
the average number of checkout lanes.
Answer by math_tutor2020(3816)   (Show Source):
You can put this solution on YOUR website!
I'll get you started on the first part of the question.
𝜇 = E(X) = 6.66 is the expected value.
There were, on average, this many checkout lanes in a supermarket that was surveyed.
Work Shown:
| X | P(X) | X*P(X) | | 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 | | Sum | 6.66 |
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