MATH 1530

Unit 2 Probability

4.2 Binomial Distributions

Binomial Probability Distributions and Random Variables

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.

1. Are the following binomial? If not, explain why not.
   1. Tossing a coin 20 times to see how many tails occur. Binomial. Fixed number of trials (20). Independent. Two categories (H, T), Probability of success remains the same.
   2. Asking 200 people if they have watched ABC news in the past week. Binomial. Fixed number of trials (200). Independent. Two categories (Y, N), Probability of success remains the same.
   3. Asking 20 people how old they are. No: More than 2 possible outcomes.
   4. Rolling a die to see if you roll a 5. Binomial. Fixed number of trials (1). Independent. Two categories (Y, N), Probability of success remains the same.
   5. Rolling a die until a 5 appears. No: Not a fixed number of trials

Probabilities in a Binomial Distribution

2. Assume that a procedure yields a binomial distribution with a trial repeated n times. Find the probability of x successes given the probability of success (p) on a given trial.

   1. n = 14 trials

      x = 4 successes

      p = .60 probability of success

      \(P(x=4)=_{14} C_{4}(.60)^{4}(.40)^{10}=1001(.6)^{4}(.4)^{10}=.0136\)

      \(q=.40=\) Probability of Failure

   2. n = 15 trials

      x = 13 successes

      p = 1/3 probability of success

      \(q=\frac{2}{3}\)

      \(P(x=13)=_{15} C_{13}\left(\frac{1}{3}\right)^{13}\left(\frac{2}{3}\right)^{2}=.0000293\)

   3. n = 23 trials

      x = 12 successes

      p = .4 probability of success

      \(q=1-0.4=0.6\)

      \(P(x=12)=_{23} C_{12}(0.4)^{12}(0.6)^{11}=.082297\)

 

3. You toss a fair coin 3 times.
   1. Create a probability distribution function (in table form) for the outcomes.

      n = 3

      p = 0.5

      q = 0.5

      Note: x is the number of heads out of the three tosses. We will need to find P(x) for all possible outcomes to create the table.

      P(x = 0) = \(_{3} C_{0}(.5)^{0}(.5)^{3}=.125\)

      P(x = 1) = \(_{3} C_{1}(.5)^{1}(.5)^{2}=.375\)

      P(x = 2) = \(_{3} C_{2}(.5)^{2}(.5)^{1}=.375\)

      P(x = 3) = \(_{3} C_{3}(.5)^{3}(.5)^{0}=.125\)

      x values	Probability P(x)
      0	.125
      1	.375
      2	.375
      3	.125

   2. What is the probability of tossing AT MOST 2 heads?

      \(P(x \leq 2)=.125+.375+.375=.875\)

   3. What is the probability of tossing AT LEAST 2 heads?

      \(P(x \geq 2)=.375+.125=.5\)

4. A quiz has two multiple choice questions each with four answer choices.
   1. Create a probability distribution function table for the number of correct answers if a student randomly selects answers.

      n = 2

      p = \(\frac{1}{4}=.25\)

      q = \(\frac{3}{4}=.75\)

      x = number correct

      x values	Probability P(x)
      0	.5625
      1	.375
      2	.0625

   2. What is the probability you get NO MORE THAN 1 correct? \(P(x \leq 1)=.9375\)
   3. What is the probability you get AT LEAST 1 correct?\(P(x \geq 1)=.4375\)

 

5. Checking Your Answers with Technology. 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

   1. Find the probability that all of the children are vaccinated. \(P(x=30)=.082\)
   2. Find the probability that exactly half of the children are vaccinated. \(P(x=15)=.00000000156\)
   3. Find the probability that at least half of the children are vaccinated. \(P(x \geq 15) \approx 1\)
   4. Find the probability that at most 25 of the children are vaccinated. \(P(x \leq 25)=.087\)
   5. Find the probability that between 25 and 30 children, inclusive, are vaccinated. \(P(25 \leq x \leq 30)=.9707\)
   6. Find the probability that between 25 and 30 children are vaccinated. \(P(26 \leq x \leq 29)=.8307\)
   7. 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.

   8. Would 24 children out of the 30 being vaccinated be unusual? Why or why not?

      Yes, it is unusual.

      \(P(x=24)=.021\)

      \(.021<.05\)

 

6. Using Technology: In a region, 80% of the population has red hair. If 30 people are randomly selected, find the following probabilities and identify the unusual events. Use a significance level of 0.05.

   By hand gets complicated: \(P(x \geq 27)=P(x=27)+P(x=28)+P(x=29)+P(x=30)\)

   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




7. A pharmaceutical company receives large shipments of aspirin tablets. The acceptable sampling plan is randomly select and test 21 tablets, then accept the whole batch if there is no more than 1 tablet in the sample that does not meet the required specifications. If a particular shipment of thousands of aspirin actually has a 3% rate of defects, what is the probability this whole shipment will be accepted? What is the probability it will be rejected?

   \(n=21 ; p=.03 ; q=.97\)

   Accepted: \(x=0 \text { or } x=1 ; P(x \leq 1)=.8701\)

   Rejected: \(P(x>1)=P(x \geq 2)=.1299 \text{ OR 1 - P(accepted)}=1-.8701=.1299 \)

Mean, Variance and Standard Deviation of a Binomial Random Variable

Binomial Distribution Center and Spread Formulas:

   Mean: \(\mu=n \cdot p\)

   Variance: \(\sigma^{2}=n \cdot p \cdot q\)

   Standard Deviation: \(\sigma=\sqrt{n \cdot p \cdot q}\)

Where

   n = number of fixed trials

   p = probability of success in one of the n trials (can be given as a percentage OR as a probability)

   q = probability of failure in one of the n trials, q = 1 - p

8. McDonald’s has a 95% recognition rate. A special focus group consists of 12 randomly selected adults.
   1. 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: \(\sigma=\sqrt{\text { variance }}\) \(= \sqrt{0.57} \approx 0.75498\)

   2. 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.

9. 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,
   1. 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.

   2. What is the probability that the drug will be effective for less than half of them? n=15, p=0.65

      \(P(7 \text { or fewer })=P(x \leq 7)=0.11323\)