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F inding m and/or s if one/two probabilities are known

P(Z<-1.645) = 0.95. P(Z<-0.842) = 0.2. 5%. 5%. 20%. 20%. Z = 1.645. Z = -0.842. F inding m and/or s if one/two probabilities are known 50 marks is the pass mark 20% of students fail 85 marks is a grade A 5% of students get A‘s.

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F inding m and/or s if one/two probabilities are known

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  1. P(Z<-1.645) = 0.95 P(Z<-0.842) = 0.2 5% 5% 20% 20% Z = 1.645 Z = -0.842 Finding mand/or sif one/two probabilities are known 50 marks is the pass mark 20% of students fail 85 marks is a grade A 5% of students get A‘s Remember: draw a diagram of the original normal data followed by a diagram of the standardised data If 5% get more than 85 then 95% get less than 85 From invnormal (0.95) F(1.645) = 0.95 So the cumulative probability up to Z = 1.645 is 0.95 If 20% get less than 50 then From the invnormal (0.2) F(-0.842) = 0.2 So the cumulative probability up to Z = -0.842 is 0.2 50 85

  2. Using Z = If x = 50 marks -0.842 = -0.842s + m = 50 If x = 85 marks 1.645 = • 1.645s + m = 85 Solving simultaneously m = 62 and s = 14 marks

  3. and We have N.B. is negative So, and e.g. 2 Find m and sif and and Solution: Using InvNormal(0.15) and InvNormal(0.75),

  4. and e.g. and We must solve simultaneously: We can change the signs in the 1st equation and add: Adding: Substitute into either equation to find m:

  5. and So, < = × P ( Z z ) 0 05 1 e.g.3 The lengths of a batch of rods follows a Normal distribution. 10%of the rods are longer than 101 cm. and 5% are shorter than 95 cm. Find the mean length and standard deviation. Solution: Let X be the random variable “length of rod (cm)” So,

  6. e.g.3 The lengths of a batch of rods follows a Normal distribution. 10% of the rods are longer than 101 cm. and 5% are shorter than 95 cm. Find the mean length and standard deviation. and Solving simultaneously: Adding: We can change the signs in the 2nd equation and add: Substitute into either equation to find m:

  7. Exercise 1. Find the values of m and s if and and 2. A large sample of light bulbs from a factory were tested and found to have a life-time which followed a Normal distribution. 25% of the bulbs failed in less than 2000 hours and 15% lasted more than 2200 hours. Find the mean and standard deviation of the distribution.

  8. and We have So, and 1. and and Solution: Using the Percentage Points of the Normal Distribution table,

  9. and e.g. and Solving simultaneously: We can change the signs in the 1st equation and add: Adding: Substitute into either equation to find m:

  10. So, We know that and So, and 2. A large sample of light bulbs from a factory were tested and found to have a life-time which followed a Normal distribution. 25% of the bulbs failed in less than 2000 hours and 15% lasted more than 2200 hours. Find the mean and standard deviation of the distribution. Solution: Let X be the random variable “life of bulb (hrs)”

  11. So, and Adding: z1 is negative Using the InverseNormal(0.25) and InverseNormal(0.85),

  12. 3 4 2 1 4 3 2 1 0 – – – – 1 3 4 2 means where So, Finding the mean or S.D e.g.1 Find the values of m in the following: and 80% 20% Solution: – 4 – 3 – 2 – 1 0 1 2 3 4 50 Remember: draw a diagram of the original normal data followed by a diagram of the standardised data Using the InverseNormal(0.8), F(0.8416) = 0.8 It doesn’t matter whether we find the z value first or use the standardising formula first.

  13. We had The mean is clearly less than 50 so the answer is reasonable. Tip: It’s easy to make a mistake and add instead of subtract or vice versa so check that your answer is reasonable by comparing with the information in the question.

  14. means where So, Find the values of the unknown in the following: and 55 70 Solution: Remember: draw a diagram of the original normal data followed by a diagram of the standardised data Using the InvNormal(0.95), F(1.6449) = 0.95

  15. Exercise 1. Find the values of the unknowns in the following: (a) and (b) and

  16. where So, Solution: (a) and Using the InvNormal(0.84),

  17. where So, Solution: (b) and Using InvNormal(0.97),

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