The document provides an overview of probability concepts, including definitions and types of distributions such as binomial, normal, and Poisson distributions. It discusses properties, exercises, and equations related to these distributions, along with practical examples. Additionally, it includes exercises related to calculating probabilities based on different scenarios and distributions.

1. Definition of probability, Binomial distribution,
Normal distribution, Poisson’s distribution,
properties – problems

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3. Basic terminology
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4. Basic terminology
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5. Basic terminology
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6. Mathematical / classical probability equation
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7. Mathematical / classical probability equation
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Exercise

9. Exercise
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10. Exercise Probability Distribution for tablet attributes
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11. Exercise
Condition 2. If specked and chipped events are not mutually exclusive
equation is … P(A or B) = P(A) + P(B) - P(A and B)
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12. Exercise
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13. Exercise Probability Distribution where attributes are not all mutually
exclusive.
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14. Exercise
The multiplicative law of probability when are not mutually exclusive
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16. 1
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17. with discrete data
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B i n o m i a l D i s t r i b u t i o n

18. Example: the result of treatment with an antibiotic is that an infection is
either cured or not cured within five days.
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B i n o m i a l D i s t r i b u t i o n
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19. 1
Probability distribution of a binomial outcome based on a
single observation.
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B i n o m i a l D i s t r i b u t i o n

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B i n o m i a l D i s t r i b u t i o n

21. expansion of the expression:
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B i n o m i a l D i s t r i b u t i o n

22. expansion of the expression:
- For, the result of treatment with an antibiotic , p = 0.75 and q =0.25
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B i n o m i a l D i s t r i b u t i o n

23. expansion of the expression:
- Further expansion of equation
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B i n o m i a l D i s t r i b u t i o n

24. expansion of the expression:
- Consider the binomial distribution with p = 0.75 and N = 4 for the
antibiotic example.
- There are five possible outcomes
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B i n o m i a l D i s t r i b u t i o n

25. expansion of the expression:
1. no patients are cured X = 0
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B i n o m i a l D i s t r i b u t i o n

26. expansion of the expression:
2. three patients are cured X =3
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B i n o m i a l D i s t r i b u t i o n

27. 2
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B i n o m i a l D i s t r i b u t i o n

28. Exercise : Calculate the probability of one success in 100 trials if p =
0.01
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B i n o m i a l D i s t r i b u t i o n

29. THE BINOMIAL DISTRIBUTION – with continuous data
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B i n o m i a l D i s t r i b u t i o n

30. THE BINOMIAL DISTRIBUTION – with continuous data
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B i n o m i a l D i s t r i b u t i o n

31. 3
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N o r m a l D i s t r i b u t i o n

32. 3
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N o r m a l D i s t r i b u t i o n

33. - As value of x increases numerically on either side of top (x= μ ), the
value of P(x) decreases rapidly
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N o r m a l D i s t r i b u t i o n

34. - The equation for the normal distribution (normal probability density)
- Where,
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N o r m a l D i s t r i b u t i o n

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N o r m a l D i s t r i b u t i o n

36. 3
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N o r m a l D i s t r i b u t i o n

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N o r m a l D i s t r i b u t i o n

38. - In this example, the histogram of tablet weights looks like a normal
curve.
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N o r m a l D i s t r i b u t i o n

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N o r m a l D i s t r i b u t i o n

40. 4
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N o r m a l D i s t r i b u t i o n

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N o r m a l D i s t r i b u t i o n

42. The standard normal probability curve has a
mean of 0 and a s.d. of 1
- The equation becomes..
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N o r m a l D i s t r i b u t i o n

43. The standard normal curve has a mean of 0 and a s.d. of 1
- a table of cumulative areas under the standard normal curve, giving
the area below Z.
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N o r m a l D i s t r i b u t i o n

45. 4
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N o r m a l D i s t r i b u t i o n

46. 4
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N o r m a l D i s t r i b u t i o n

47. 4
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N o r m a l D i s t r i b u t i o n

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N o r m a l D i s t r i b u t i o n

49. the Z transformation
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N o r m a l D i s t r i b u t i o n

50. Limits,
no tablet may differ from the average weight by more than 20%
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51. the Z transformation less than 80.96mg and fining area using the Table
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N o r m a l D i s t r i b u t i o n

52. the Z transformation less than 121.44mg and fining area using the Table
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N o r m a l D i s t r i b u t i o n

53. the Z transformation
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55. 5
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P o i s s o n D i s t r i b u t i o n

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P o i s s o n D i s t r i b u t i o n

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P o i s s o n D i s t r i b u t i o n

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P o i s s o n D i s t r i b u t i o n

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P o i s s o n D i s t r i b u t i o n

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P o i s s o n D i s t r i b u t i o n

61. 1. What is the probability that three of six patients will be cured if the
probability of a cure is 60%? Ans = 0.276
2. What is the probability of observing exactly one head in 10 tosses of
a coin? Ans = 0.0098
3. during production of a dosage form, the pharmaceutical company
normally expects to have 2% of the tablets in a batch to have less
than 95% of the labeled amount of a drug (defective). If 20 tablets are
randomly sampled from a batch, what is the probability of finding
three defective tablets? Ans = 0.0067
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62. 4. If the mean and standard deviation of the potency of a batch of
tablets are 50 and 5 mg, respectively, what proportion of the tablets
have a potency between 40 and 60 mg?
Ans =0.954
5. Serum sodium values for normal persons have a mean of 140 mEq/L
and a s.d. of 2.5. What is the probability that a person’s serum
sodium will be between 137 and 142 mEq/L?
Ans = 0.673
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63. End
Reference:
Ph’ceutical statistics- Practical and clinical
applications –Sanford Bolton
Fundamental of Statistics –S.C. Guptha
Net Source
of the topic