A new young executive is perplexed at the number of interruptions that occur due
ID: 3124359 • Letter: A
Question
A new young executive is perplexed at the number of interruptions that occur due to employee relations. She has decided to track the number of interruptions that occur during each hour of her day. Over the last month, she has determined that between 0 and 3 interruptions occur during any given hour of her day. The data is shown below. Number of Interruptions in 1 hour& Probability 0 interruption 0.5 , 1 interruptions 0.3, 2 interruptions 0.1, 3 interruptions 0.1 on average, how many interruptions should she expect per hour. What is Variance and the standard deviation of interruptions?
Explanation / Answer
Consider:
Thus,
E(x) = Expected value = mean = 0.8 [ANSWER]
*****************
Var(x) = E(x^2) - E(x)^2 = 0.96 [ANSWER, VARIANCE]
s(x) = sqrt [Var(x)] = 0.979795897 [ANSWER, STANDARD DEVIATION]
x P(x) x P(x) x^2 P(x) 0 0.5 0 0 1 0.3 0.3 0.3 2 0.1 0.2 0.4 3 0.1 0.3 0.9Related Questions
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