5. Computing probability from a frequency distribution Aa Aa Consider the follow

Question

5. Computing probability from a frequency distribution Aa Aa Consider the following frequency distribution histogram for a population that consists of N = 9 scores FREQUENCY 4 Suppose you take a random sample of one score from this set. The probability that this score is equal to 3 is p(X-3) = score is greater than 3 is p(X > 3) = The probability that this score is less than 3 is p(X 3) The probability that this 0.47 0.67 0.33 0.43 A bucket contains many tickets with different values. Let X be the outcome of ar table below shows a probability distribution table for X. aw from this bucket. The X Value 1 2 34 5 p(X) 1/9 2/9 3/9 2/9 1/9 Suppose you take a random sample of one score from this bucket. The probability that this score shows a 3 is p(X = 3) that this score shows a number greater than 3 is p(X > 3) The probability that this score shows a number less than 3 is p(X

Explanation / Answer

Solution:

From this histogram:

P(X=1) = 1/9

P(X=2) = 2/9

P(X=3) = 3/9

P(X=4) = 2/9

P(X=5) = 1/9

So P(X=3) =3/9 = 0.33

P(X<3) = P(X=1) + P(X=2) = 1/9 + 2/9 = 3/9 = 0.33

P(X>3) = P(X=4)+P(X=5) = 2/9+1/9 = 3/9 = 0.33

As given in the question probability of each value: so we are calculating

So P(X=3) =3/9 = 0.33

P(X<3) = P(X=1) + P(X=2) = 1/9 + 2/9 = 3/9 = 0.33

P(X>3) = P(X=4)+P(X=5) = 2/9+1/9 = 3/9 = 0.33

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