The following table shows the probability distribution of for the discrete random variable X= the number of traffic accidents reported in a day in a small city in the Midwest. x 0 1 2 3 P(x) 0.23 0.17 0.26 0.34 1. The probability of less than 1 accident is: a. 0.23 b. 0.40 c. 0.32 d. 0.60 2. The mean (expected value) of X is: a. 1.17 b. 1.71 c. 0.17 d. 0.71 3. The probability that X will be at least 2 is: a. 0.26 b. 0.34 c. 0.60 d. 0.66
Added by Brenda W.
Step 1
From the given table, P(X = 0) = 0.23. Step 2: The mean (expected value) of X can be calculated using the formula E(X) = Σ(x * P(x)), where x is the value of the random variable and P(x) is the probability of that value. E(X) = 0 * 0.23 + 1 * 0.17 + 2 * 0.26 + Show more…
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The following table contains the probability distribution for the number of traffic accidents daily in a small town. Complete parts (a) through (c) below. Number of accidents per day 0 0.28 1 0.33 2 0.15 3 0.09 4 0.05 5 0.07 6 0.03 a. Compute the mean number of accidents per day. b. Compute the standard deviation. c.What is the probability that there will be at least 2 accidents on a given day?
Determine whether or not the table is a valid probability distribution of a discrete random variable. Explain fully. a. \begin{tabular}{c|ccccc} & 0 & 1 & 1 & 3 & 4 \\ \hline$P(x)$ & -0.85 & 0.50 & 0.25 & 0.10 & 0.30 \end{tabular}b. \begin{tabular}{c|ccc} & 1 & 1 & 3 \\ \hline$P(x)$ & 0.325 & 0.406 & 0.164 \end{tabular} C. \begin{tabular}{c|ccccc} & 25 & 26 & 27 & 28 & 20 \\ \hline$P(x)$ & 0.13 & 0.27 & 0.28 & 0.18 & 0.14 \end{tabular}
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