2.4 Binomial Probability Distributions
Binomial Probability Distributions result from a procedure that meets all the following requirements:
- The procedure has a FIXED number of trials
- The trials are INDEPENDENT (the outcome of any individual trial does not affect the probabilities in the other trials.)
- All trial outcomes can be classified into TWO CATEGORIES (success and failure).
- The probability of success REMAINS THE SAME throughout the trials.
- In a region, 80% of the population has red hair. If 30 people are randomly selected, find the following probabilities for red hair and identify the unusual events. Use a significance level of 0.05.
Words Probability Statement Calculated Probability Unusual (Yes or No) Probability of at least 27 out of 30 \(P(x \geq 27)\)
.1227
no
Probability of less than 25 out of 30 \(P(x<25)\)
.5725
no
Probability of no more than 18 out of 30 \(P(x \leq 18)\)
.0095
yes
Probability of more than 21 out of 30 \(P(x>21)\)
.8713
no
Probability of at least 26 out of 30 \(P(x \geq 26)\)
.2552
no
Probability of 16 out of 30 \(P(x=16)\)
.0007
yes
Probability of less than 4 out of 30 \(P(x<4)\)
.0000+
yes
- In 2017, 92% of American children were vaccinated against measles. https://data.worldbank.org/indicator/sh.imm.meas
A health worker wants to gather some data to see if the vaccination rate may have changed recently. She randomly selects 30 children for her study.
n =
30 p =
.92 q =
.08 - Find the probability that all of the children are vaccinated.
\(P(x=30)=.082\) - Find the probability that exactly half of the children are vaccinated.
\(P(x=15)=.00000000156\) - Find the probability that at least half of the children are vaccinated.
\(P(x \geq 15) \approx 1\) - Find the probability that at most 25 of the children are vaccinated.
\(P(x \leq 25)=.087\) - Find the probability that between 25 and 30 children, inclusive, are vaccinated.
\(P(25 \leq x \leq 30)=.9707\) - Find the probability that between 25 and 30 children are vaccinated.
\(P(26 \leq x \leq 29)=.8307\) - If at most 25 of the children are vaccinated, does it appear that the 92% vaccination rate is wrong?
\(.087>.05\)
Not unusual, so probably not wrong.
- Would 24 children out of the 30 being vaccinated be unusual? Why or why not?
Yes, it is unusual.
\(P(x=24)=.021\)
\(.021<.05\)
- Find the probability that all of the children are vaccinated.
- McDonald’s has a 95% recognition rate. A special focus group consists of 12 randomly selected adults.
- For such a group, find the mean, variance, and standard deviation.
Mean: \(\mu=n \cdot p\)
\(=12 \cdot(0.95)=11.4\) Variance: \(\sigma^{2}=n \cdot p \cdot q\)
\(=(12)(0.95)(0.05)=0.57\) Standard Deviation: \(\sigma=\sqrt{\text { variance }}\)
\(= \sqrt{0.57} \approx 0.75498\) - Use the range rule of thumb to find the minimum and maximum usual number of people who would recognize McDonald’s.
Minimum: \(\mu-2 \sigma = \)
\(11.4-2(.755)=9.89\) Maximum: \(\mu+2 \sigma=\)
\(11.4+2(.755)=12.91\) According to the range rule of thumb, it would not be unusual for 10, 11 or 12 people out 12 to recognize McDonalds. It would be unusual for 9 or fewer out of 12.
- For such a group, find the mean, variance, and standard deviation.
- Suppose that Bayanisthol, a new drug, is effective for 65% of the participants in clinical trials. If a group of fifteen patients take this new drug,
- What is the expected number of patients for whom the drug will be effective?
We expect the mean. Expected number = mean = \(\mu=n \cdot p=15 \cdot(0.65)=9.75\) We would expect 9.75 patients to have positive effects.
- What is the probability that the drug will be effective for less than half of them?
n=15, p=0.65 \(P(7 \text { or fewer })=P(x \leq 7)=0.11323\)
- What is the expected number of patients for whom the drug will be effective?