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anonymous
 5 years ago
manager of a retail store that employees 80 workers tracked the number of times each employee was late to work during a 1month period. The results are summarized in the table below.
Days Late: 0 1 2 3 4 5 6
# Employ:47 13 8 6 4 1 1
Find the expected number of days an employee will be late in a 1month period.
[Give your answer with 3 decimal places.]
anonymous
 5 years ago
manager of a retail store that employees 80 workers tracked the number of times each employee was late to work during a 1month period. The results are summarized in the table below. Days Late: 0 1 2 3 4 5 6 # Employ:47 13 8 6 4 1 1 Find the expected number of days an employee will be late in a 1month period. [Give your answer with 3 decimal places.]

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The expected value of an observable X (here, 'days late'), is given by\[E(X)=\sum_{i=1}^{n}x_ip(x_i)\]where\[x_i\]is the value of the observable and \[p_(x_i)\]is the probability of observing it. Here, you have the x_i's being 'days late at step i' and p(x_i) the probability of observing it. x_0 = 0 and p(x_0=0)=47/80 x_1 = 1 and p(x_1=1)=13/80 x_2 = 2 and p(x_2=2)=8/80 x_3 = 3 and p(x_3=3)=6/80 x_4 = 4 and p(x_4=4)=4/80 x_5 = 5 and p(x_5=5)=1/80 x_6 = 6 and p(x_6=6)=1/80 From the definition, \[E(X)\]\[=0.\frac{47}{80}+1.\frac{13}{80}+2.\frac{8}{80}+3.\frac{6}{80}+4.\frac{4}{80}+5.\frac{1}{80}+6.\frac{1}{80}\]\[=\frac{74}{80}\]\[=0.925\]days. Please hit my "Become a fan" button/link (next to my name) if this helps :)
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