Question

In a recent fundraising campaign, Rise Over Run received 20 checks of various amounts to donate to their favorite charities. Olive uses this spinner to determine which charity will receive each check.
About how many of the 20 checks should arise Over Run expect to go to their TreeHuggers Campaign ?

Answer 1

Answer:

6 checks

Step-by-step explanation:

Answer 2

The expected number of checks out of 20 checks that should arise Over Run expect to go to their TreeHuggers Campaign is given by: Option D: 6 checks.

How to find that a given condition can be modeled by binomial distribution?

Binomial distributions consists of n independent Bernoulli trials.

Bernoulli trials are those trials which end up randomly either on success (with probability p) or on failures( with probability 1- p = q (say))

Suppose we have random variable X pertaining binomial distribution with parameters n and p, then it is written as

$X \sim B(n,p)$

The probability that out of n trials, there'd be x successes is given by

$P(X =x) = \: ^nC_xp^x(1-p)^{n-x}$

The expected value and variance of X are:

$E(X) = np\\ Var(X) = np(1-p)$

For this case, we're specified that:

The spinner has 10 parts.

• 3 belong to Treehuggers
• 2 belong to Ameriteach
• 5 belong to Paws of Peace

Thus, the probability of getting the spinner fall on treehugger is 3/10 (three favorable parts of spinner to ten total parts, all parts assumingly equally probable and spinner assumingly not falling in the mid of two parts).

20 checks, all will be accompanied with 20 spins, all independent of each other.

Let X = total number of times the spinner falls on Treehuggers

And let Success for a spin = It landing on Treehugger

Failure for a spin = It not landing on Treehugger

• Probability of success = p = 3/10 = 0.3
• Probability of faillure = 1-p = 7/10 = 0.7

n = 20, so we get:

$X \sim B(n=20,p=0.3)$

The expected number of checks out of 20 checks that should arise Over Run expect to go to their TreeHuggers Campaign = expected number of successes in those 20 spins =  the expected number of values of X.

Since X follows a binomial distribution, we get:

$E(X) = np = 20 \times 0.3 = 6$

Thus, the expected number of checks out of 20 checks that should arise Over Run expect to go to their TreeHuggers Campaign is given by: Option D: 6 checks.

Learn more about binomial distribution here:

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