Standardized Test Practice
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Use the Binomial Theorem.
Add the probabilities that 7 and 8 students do not buy compact discs.
Recall the formula for the mean of a binomial distribution.
| Number of Successes | Probability |
|---|---|
| 8 | ≈ 0.4703 |
| 7 | ≈ 0.3721 |
| 6 | ≈ 0.1288 |
| 5 | ≈ 0.0255 |
| 4 | ≈ 0.0031 |
| 3 | ≈ 0.0002 |
| 2 | ≈ 0.0000 |
| 1 | ≈ 0.0000 |
| 0 | ≈ 0.0000 |
≈ 0.8424
7
We are asked to determine the probabilities associated with the number of students that do not buy compact discs. First, let's look at the results of the recent survey conducted on all high school students. We will convert the percentages to decimal notation for convenience.
| Students Who Do Not Buy Compact Discs | Students Who Buy Compact Discs | |
|---|---|---|
| Percentages | 91 % | 9 % |
| Decimals | 0.91 | 0.09 |
Asking students if they buy compact discs is a binomial experiment. In our case, a success is the event that a student does not buy compact discs. A failure is if the student buys compact discs. The probabilities of success and failure are p= 0.91 and q= 0.09, respectively.
| Type Of Event | Description | Probability |
|---|---|---|
| Success | Student does not buy compact discs. | p= 0.91 |
| Failure | Student buys compact discs. | q= 0.09 |
We can find the distribution and the probabilities associated with the survey using a probability binomial ( p+ q)^n. We know that 8 students have been surveyed, so we will substitute n= 8, p= 0.91, and q= 0.09 into the binomial. ( p+ q)^n ⇔ ( 0.91+ 0.09)^8 To find the probabilities, we will expand the binomial using the Binomial Theorem. Let s represent students who do not buy compact discs and f represent students who buy compact discs.
Let's write down all the probabilities in a table.
| Number of Successes | Number of Failures | Probability | Simplified |
|---|---|---|---|
| 8 | 0 | 0.91^8 | ≈ 0.4703 |
| 7 | 1 | 8* 0.91^7* 0.09^1 | ≈ 0.3721 |
| 6 | 2 | 28* 0.91^6* 0.09^2 | ≈ 0.1288 |
| 5 | 3 | 56* 0.91^5* 0.09^3 | ≈ 0.0255 |
| 4 | 4 | 70* 0.91^4* 0.09^4 | ≈ 0.0031 |
| 3 | 5 | 56* 0.91^3* 0.09^5 | ≈ 0.0002 |
| 2 | 6 | 28* 0.91^2* 0.09^6 | ≈ 0.0000 |
| 1 | 7 | 8* 0.91^1* 0.09^7 | ≈ 0.0000 |
| 0 | 8 | 0.09^8 | ≈ 0.0000 |
The probability that at least 7 of the 8 students do not buy compact discs is the sum of the probabilities of the events where 7 or 8 students do not buy compact discs. In Part A, we determined that a success is when a student does not buy compact discs. To find the desired probabilities, let's look at the table from Part A.
| Number of Successes | Probability |
|---|---|
| 8 | ≈ 0.4703 |
| 7 | ≈ 0.3721 |
| 6 | ≈ 0.1288 |
| 5 | ≈ 0.0255 |
| 4 | ≈ 0.0031 |
| 3 | ≈ 0.0002 |
| 2 | ≈ 0.0000 |
| 1 | ≈ 0.0000 |
| 0 | ≈ 0.0000 |
We are looking for the sum of the approximate values of the probabilities from the first two rows. Let's calculate it! 0.4703+0.3721=0.8424 The probability that at least 7 of the 8 students do not buy compact discs is equal to about 0.8424.
In Part A, we concluded that the survey is a binomial experiment. A success is when the student does not buy compact discs. The expected number of students who do not buy compact discs is the expected number of successes. To find this, we will first find the mean of the binomial distribution.
|
Mean of a Binomial Distribution |
|
The mean μ is given by μ=n* p, where n is the number of trials and p is the probability of success. |
We previously found that the probability of success is p= 0.91 and the number of trials, which in this case is the number of students surveyed, is n= 8. Let's substitute these values into the equation for μ.
The mean is 7.28. To find the expected number of successes, we need to round the mean to the nearest integer, because a fraction of success makes no sense. Therefore, the expected number of successes is 7 .