![Binomial distribution (6)
• The process could be repeated for any number
of trials or any number of patients but to simply
process can use the binomial formula
• ! Refers to factorial
– Example: 8! = 8x7x6x5x4x3x2x1
10
n combination x
Gives the number of
times that particular
combination of events
can occur out of n
independent trials [nCx]](https://image.slidesharecdn.com/probabilitydistributionsforsharing2024-240229174453-2f790d4c/85/Probability-distributions-FOR-SHARING-2024-pptx-10-320.jpg)
![Binomial distribution (6)
• Suppose 10 men with prostate cancer are
chosen
– The probability that all 10 men survive n=10,
X=10
– Probability that exactly 10 men survive?
– Probability that at least 2 men survive? P(X≥2) =
P(2)+P(3)……+P(10) = 1- P(X<2) = 1-[P(1) + P(0)]
= 0.999995801
– Probability that at most 3 men survive? P(X≤3) =
P(0)+P(1)+P(2)+P(3) = 0.00086436 11](https://image.slidesharecdn.com/probabilitydistributionsforsharing2024-240229174453-2f790d4c/85/Probability-distributions-FOR-SHARING-2024-pptx-11-320.jpg)
Probability distributions FOR SHARING 2024.pptx
- 1. 1 Random variables and probability distributions
- 2. 2 Probability distributions • Is summary of the probabilities of occurrence of the different levels of a random variable. • A random variable is a variable in a study where subjects have been selected randomly • A variable is a characteristic that varies from subject to subject
- 3. 3 Probability distributions 2 • If had heights (X) of 10 women selected randomly: 152, 160, 165, 158, 155, 153, 168, 165, 163, 156. – The height (X) is a random variable. – Can determine probability that it has any given value or range of values. – E.g. probability that X is less than 160cm = 5 / 10. or probability that height is 155cm = 1 / 10
- 4. 4 Probability distributions 3 • To create probability distributions the determined probabilities can be plotted against characteristic (X) • Can use theoretical distribution to fit distribution of variable of interest. • Examples of theoretical distributions are: - Binomial & Poisson distributions – deal with discrete random variables (take on only integers) - Normal (Gaussian) – deals with continuous random variables
- 5. Binomial distribution • Is theoretical distribution that is applicable to events that have binary outcomes • Two possible outcomes denoted as A and B • P(A) = Π P(B) = 1- Π • Probability stays same each time event occurs • Outcome is independent from one trial to another 5
- 6. Binomial distribution (2) • Gives probability that specified outcome occurs in a given number of independent trials • If an experiment involving this event is repeated n times and the outcome is independent from one trial to another, what is the probability that outcome A occurs exactly X times? • Or equivalently, what proportion of the n outcomes will be A? 6
- 7. Binomial distribution (3) • Assume population of men with localized prostate tumor and pretreatment PSA < 10 studied, & probability of 5-year survival = 0.8. – S represent event of 5-year survival; π = P(S) = 0.8 – D represent death before 5 years; 1 – π = P(D)= 0.2 – Consider group of n = 2 men with a localized prostate tumor and pretreatment PSA < 10. 7
- 8. Binomial distribution (4) • What is probability that exactly two men live 5 years? – P(Survival for patient 1 and survival patient 2) – Apply multiplicative rule since the events are independent survival of one patient does not affect survival of another – P(S1 and S2) = P(S) X P(S) = 0.8 x 0.8 = 0.64 8
- 9. Binomial distribution (5) • What is probability that exactly one man lives 5 years? – P(S1 and D2) or P(D1 and S2) – Both multiplicative rule and addition rule apply – P(S)X(PD) + P(D)XP(S) = (0.8x0.2)+(0.2x0.8) =0.32 • What is probability that none lives 5 years? – P(D1 and D2) = P(D) x P(D) = 0.2 x 0.2 = 0.04 9
- 10. Binomial distribution (6) • The process could be repeated for any number of trials or any number of patients but to simply process can use the binomial formula • ! Refers to factorial – Example: 8! = 8x7x6x5x4x3x2x1 10 n combination x Gives the number of times that particular combination of events can occur out of n independent trials [nCx]
- 11. Binomial distribution (6) • Suppose 10 men with prostate cancer are chosen – The probability that all 10 men survive n=10, X=10 – Probability that exactly 10 men survive? – Probability that at least 2 men survive? P(X≥2) = P(2)+P(3)……+P(10) = 1- P(X<2) = 1-[P(1) + P(0)] = 0.999995801 – Probability that at most 3 men survive? P(X≤3) = P(0)+P(1)+P(2)+P(3) = 0.00086436 11
- 12. Binomial distribution (7) • The mean of a binomial distribution is nπ – Thus for above distribution of prostate cancer patients with n=10 mean = 10x0.8 = 8 – The standard deviation is square root of nπ(1-π) (10x0.8x0.2)^0.5=1.265 – Thus the parameters of binomial distribution are n and π because they are the only two parameters needed to completely describe the binomial distribution 12
- 13. Binomial distribution (8) • Studies of binary variables often report proportions rather than the number of subjects with a certain characteristic – Even in such scenarios n and π are the parameters required – Since a proportion is got by dividing X by n the mean of a proportion becomes π – The standard deviation is 13
- 14. Poisson distribution • Is a discrete distribution when outcome is number of times an event occurs – Used to determine the probability of rare events – Gives probability that outcome occurs specified number of times when the number of trials is large and the probability of any one occurrence is small • used to plan number of beds needed in ICU of a hospital • number of ambulances needed on call • model number of cells in a given volume of fluid • number of bacterial colonies growing in a certain amount of medium, or the emission of radioactive particles from a specified amount of radioactive material 14
- 15. Poisson distribution (2) • Consider random variable representing number of times event occurs in given time or space interval. – probability of exactly X occurrences is given by the formula – λ is value of both the mean and the variance and it is the only parameter of the Poisson distribution 15
- 16. Poisson distribution (3) • Consider random variable representing number of times event occurs in given time or space interval. – probability of exactly X occurrences is given by the formula – λ is value of both the mean and the variance and it is the only parameter of the Poisson distribution 16
- 17. Poisson distribution (4) • Suppose hospitalizations for patients with medical and surgical treatment follows Poisson distribution – Poisson is applicable because the chance that patient goes into hospital is small – Can be assumed to be independent from patient to patient – After 11 years, the 390 patients randomized to medical group were admitted total 1256 times (mean = ?) – Patients randomized to surgical group were admitted 1487 times (mean = ?) 17
- 18. Poisson distribution (5) • Probability of exactly 0 hospitalizations in surgical group ? – λ = 3.22 – P(0) = 0.03996 • Probability of exactly 2 hospitalizations in medical group? 18
- 19. 19 Normal distribution • Distribution for continuous random variables • Is smooth bell shaped curve, symmetrical about mean of the distribution (μ). Standard deviation of distribution is symbolized by σ. – Since it is a probability distribution, the area under the curve is = 1. – Half area is on left of the mean, half on right.
- 20. 20 Normal distribution • Approximately 68% of the data lies within 1 SD of the mean • Approximately 95% of the data lies within 2 SD of the mean • Approximately 99.7% of the data lies within 3 SD of the mean • This is known as the empirical rule.
- 21. 21 Normal distribution …continued • Can be transformed to standard normal distribution (z) which has mean = 0, standard deviation = 1. • Transformation is made by: z = X is the observation that you wish to transform µ is the mean of the characteristic in the population σ is standard deviation of characteristic in population X
- 22. 22 Normal distribution …continued Example: Supposing SBP in normal healthy individuals is normally distributed with μ = 120 mmHg, σ = 10mmHg. To get proportion of subjects with BP above 130 mmHg Z = – Draw the distribution to see which area is of interest • Read off the values from the z-table attached at end of these slides 1 10 120 130
- 23. 23 Normal distribution …continued • Proportion with BP between 110 & 130 mmHg • What proportion of subjects have BP less than 100 or above 140mmHg? • What proportion have BP less than 90mmHg or above 140mmHg? 1 10 120 130 1 10 120 110 2 1 andZ Z
- 24. 24 Normal distribution …continued • Proportion of subjects with BP above 133? • Value that divides distribution into lower 2.5 and upper 97.5% • What BP would divide the distribution into: – lower 5% and upper 95%? – Upper 10% and lower 90% 4 . 100 120 6 . 19 10 120 96 . 1 X X X
- 25. 25 Normal distribution …continued • What BP would divide the distribution into: – lower 20% and upper 80%? – The specific z-value is not appearing in the table therefore calculate using the closest values to get it – Look at area in one tail and you see there is area of 0.212 and area of 0.198 between which we shall find the 0.2 corresponding to question of interest
- 26. Finding z-value when missing from table Z area 0.80 0.212 ? 0.200 0.85 0.198 • Therefore difference of 0.02 corresponds to (0.002/0.014)*0.05 = 0.007 • 0.85-0.007 = • Difference in area 0.212- 0.198 = 0.014 corresponds to difference of (0.85-0.80) 0.05 in z • Therefore difference of 0.012 corresponds to • (0.012/0.014)*0.05 = 0.043 • Z corresponding to 0.2 = 0.8+0.043=0.843 26
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