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The Magazine Mass Marketing Company has received 14

entries in its latest sweepstakes. They know that the probability of receiving a magazine subscription order with an entry form is 0.6
. What is the probability that more than 4
of the entry forms will include an order? Round your answer to four decimal places.

User Kebs
by
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1 Answer

6 votes

Answer:

0.9825

Explanation:

The Magazine Mass Marketing Company has received 14 entries in its latest sweepstakes. They know that the probability of receiving a magazine subscription order with an entry form is 0.6.

To calculate the probability that more than 4 of the entry forms will include an order, we can model the given scenario as a binomial distribution.

Binomial distribution


\large\boxed{X\sim \text{B}(n,p)}

where:

  • X is the random variable that represents the number of successes.
  • n is the fixed number of independent trials.
  • p is the probability of success in each trial.

Given the probability of receiving a magazine subscription order with an entry form is 0.6, and the number of entries is 14:


  • p = 0.6

  • n = 14

Therefore:


X\sim \text{B}(14,0.6)

where the random variable X represents the number of entry forms that include a magazine subscription order.

To find the probability that more than 4 of the entry forms include a magazine subscription order, we need to find P(X ≥ 5).

The complement rule of probability states that the probability of an event occurring is equal to one minus the probability of that event not occurring. Therefore:


\text{P}(X \geq 5) = 1 - \text{P}(X \leq 4)

We can use a calculator to calculate P(X ≤ 4) using the binomial cumulative distribution function (cdf). Note that the binomial cdf on a calculator will give you the sum of all the binomial probabilities for values of your random variable less than or equal to a given number (i.e. P(X ≤ x)).

Inputting the values of n = 14, p = 0.6 and x = 4 into the binomial cdf we get:


\text{P}(X \leq 4) = 0.0175095414...

Therefore:


\begin{aligned} \text{P}(X \geq 5) &= 1 - \text{P}(X \leq 4)\\&= 1 - 0.0175095414...\\&= 0.982490458...\\&=0.9825\;(4\;\sf d.p.)\end{aligned}

So the probability that more than 4 of the entry forms will include a magazine subscription order is 0.9825 (4 d.p.).

User Jfbarrois
by
8.3k points