A spinner has four sectors, labeled 1, 2, 3, and 4, respectively. Each spin resu

Question

A spinner has four sectors, labeled 1, 2, 3, and 4, respectively. Each spin results in one of the four sectors being selected. The spinner in spun 120 times, and the number of 1's is denoted by f_1, the number of 2's is denoted by j, the number of 3'* i* denoted by f_2, and the number of 3's is denoted by f_3, and the number of 4's is denoted by f_4. Let p be the probability that (f_1 - 30)^2/30 + (f_2 - 30)^2/30 + (f_2 - 30)^2/30 + (f_4 - 30)^/30 is greater than 6.25. Given that the four outcomes on the spinner are equally likely, which of the following is closest to the value of p?

Explanation / Answer

This is then a chi-squared distribution with

df = k - 1 = 4 - 1 = 3

where k = 4 groups.

Hence,

P(chi^2 > 6.25) = 0.100060833 = 0.1 [ANSWER, A]

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