binomial distribution

Question

A fair 5-sided spinner has sides numbered 1, 2, 3, 4 and 5 The spinner is spun once and the score of the side it lands on is recorded.

(a) Write down the name of the distribution that can be used to model the score of the side it lands on. (1)

The spinner is spun 28 times. The random variable X represents the number of times the spinner lands on 2

(b)

(i) Find the probability that the spinner lands on 2 at least 7 times.

(ii) Find P(4 ⩽ X < 8)

Answer 1

a) We would model a single spin with the discrete uniform distribution.

As it is stated the spinner is fair, and there are 5 possible spins, P(1) = P(2) = P(3) = P(4) = P(5) = 1/5

b) As P(2) = 1/5 for each spin, we can model this with the binomial distribution.

Let x be the number of spins of 2 in 28 spins.

P(x) = C(28,x)(1/5)^x(4/5)^n-x

Statistical calculators along with Excel have functions that do this calculation so that all you have to provide is n, x, and p (trials, successes, and probability of a success in each triial) along with a flag allowing to select either the pmf, the probability that there is a particular value of x, or the cumulative distribution (the probability of x or less successes). Also, we get the probability greater than by calculating 1 - the probability less than or equal to - the cumulative distribution).

P(x >= 7) = 1 - P(x <= 6) = 1 - binom.dist(6,28,1/5,1) - the 1 indicates cumulative = 0.32155649687205

Note that E(x) = np = 28 * .2 = 5.6, so we should expect a value less than .5

P(4 <= x <= 8) = P(x <= 8) - P(x <= 3) = binom.dist(8,28,1/5,1) - binom.dist(3,28,1/5,1) = 0.749782670271184 -

we would expect a fairly large value as these values approach being the values within 1 standard deviation of the mean.

Answer 2

a) binomial. either lands on number (success) or doesn't (failure); probability lands on any of the five numbers is constant at 0.2. spins are independent, there are a fix number of spins (28).

probability lands on 2 x times out of 28: (28 x)(0.2)^x(0.8)^(28-x).

b) probability lands on 2 at least 7 times = 1 - [P(lands on 2 6 times) - P(lands on 2 5 times) - ... - P(lands on 2 1 time) - P(lands on 2 zero times)]

probability lands on 2 between 4, 5, 6 or 7 times = P(lands on 2 4 times) + P(lands on 2 5 times) + P(lands on 2 6 times) + P(lands on 2 7 times)