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A fair die is rolled 72 times and the percentage of 6s is recorded. What is the probability that at most 10% of the rolls are 6s

User Naumcho
by
7.5k points

1 Answer

6 votes

Answer:

P (
\hat p ≤ 0.10)

Explanation:

The probability in terms of statistics for this given problem can be written as follows.

Let consider X to the random variable that represents the number of 6's in 7 throws of a dice, then:

X
\simBin ( n = 72, p = 0.167)

E(X) = np

E(X) = 72× 0.167

E(X) = 12.024

E(X)
\simeq 12

p+q =1

q = 1 - p

q = 1 - 0.167

q = 0.833

V(X) = npq

V(X) = 72 × 0.167 × 0.833

V(X) = 10.02

V(X)
\simeq 10

∴ X
\sim N (
\mu = 12, \sigma^2 =10)


\hat p = (X)/(n) \sim N ( p, (pq)/(n))

where p = 0.167 and
(pq)/(n) =
(0.167 * 0.833)/(72) = 0.00193

∴ P(at most 10% of rolls are 6's)

i.e

P (
\hat p ≤ 0.10)

User Arune
by
7.8k points

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