Normal distribution<br />Unit 8 strand 1<br />
It has the following features:<br /><ul><li>bell-shaped
symmetrical about the mean
it extends from –infinity to + infinity
The area under the curve is 1</li></li></ul><li>Approximately 95% of the distribution lies between 2 SDs of the mean<br />
Approximately 99.9% of the distribution lies between 3 SDs of the mean<br />
The probability that X  lies between a and b is written as:<br />P(a<X<b).<br />To find the probability we need to find th...
The probability that X  lies between a and b is written as:<br />P(a<X<b).<br />To find the probability we need to find th...
So, the probability of Z < 0.16 is 0.5636.<br />
Find the probability of Z < 0.429<br />
Now try these<br />Draw sketches to illustrate your answers<br />If Z ~N (0, 1), find<br />1. P (Z <0.87) 	2. P (Z > 0.87)...
Example<br />Find <br />P (Z < 0.411) <br />P (Z > - 0.411) <br />P (Z > 0.411) <br />P (Z < - 0.411)<br />Solution<br />P...
Now try these<br />1. P (Z > - 0.314)	<br />2. P (Z < - 0.314)	<br />3. P (Z > 0.111)	<br />4. P (Z > - 0.111)<br />
What if we don’t have a distribution where the mean is 0 and the standard deviation is 1?<br />
Standardising a normal distribution<br />To standardise X where X ~ N(μ, σ²)<br />subtract the mean and then divide by the...
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Normal distribution

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Normal distribution

  1. 1. Normal distribution<br />Unit 8 strand 1<br />
  2. 2.
  3. 3. It has the following features:<br /><ul><li>bell-shaped
  4. 4. symmetrical about the mean
  5. 5. it extends from –infinity to + infinity
  6. 6. The area under the curve is 1</li></li></ul><li>Approximately 95% of the distribution lies between 2 SDs of the mean<br />
  7. 7. Approximately 99.9% of the distribution lies between 3 SDs of the mean<br />
  8. 8. The probability that X lies between a and b is written as:<br />P(a<X<b).<br />To find the probability we need to find the area under the normal curvebetween a and b.<br />We can integrate or use tables.<br />
  9. 9. The probability that X lies between a and b is written as:<br />P(a<X<b).<br />To find the probability we need to find the area under the normal curvebetween a and b.<br />We can integrate or use tables.<br />To simplify the tables for all possible values of mean and s² the variable X is standardised so that the mean is zero and the standard deviation is 1. The standardised normal variable is Z and Z (0, 1)<br />
  10. 10.
  11. 11. So, the probability of Z < 0.16 is 0.5636.<br />
  12. 12. Find the probability of Z < 0.429<br />
  13. 13. Now try these<br />Draw sketches to illustrate your answers<br />If Z ~N (0, 1), find<br />1. P (Z <0.87) 2. P (Z > 0.87) 3. P (Z < 0.544) 4. P (Z > 0.544)<br />
  14. 14.
  15. 15.
  16. 16. Example<br />Find <br />P (Z < 0.411) <br />P (Z > - 0.411) <br />P (Z > 0.411) <br />P (Z < - 0.411)<br />Solution<br />P(Z < 0.411) = [from tables 6591 + 4] = 0.6595<br />P(Z > - 0.411) = P(Z < 0.411) = 0.6595 (from (above))<br />P(Z > 0.411) = 1 – Φ(0.411) = 1 – 0.6595 = 0.3405<br />P(Z < - 0.411) = P(Z > 0.411) = 0.3405<br />
  17. 17. Now try these<br />1. P (Z > - 0.314) <br />2. P (Z < - 0.314) <br />3. P (Z > 0.111) <br />4. P (Z > - 0.111)<br />
  18. 18. What if we don’t have a distribution where the mean is 0 and the standard deviation is 1?<br />
  19. 19. Standardising a normal distribution<br />To standardise X where X ~ N(μ, σ²)<br />subtract the mean and then divide by the standard deviation:<br /> where Z ~ N(0,1)<br />
  20. 20.
  21. 21.
  22. 22.
  23. 23.
  24. 24. Task 3<br />A forensic anthropologist has asked your advice. She was investigating the lifespan of insects on a human corpse. The mean lifespan for one insect is 144 days and the standard deviation is 16 days. Find the probability that one insect will live less than 140 days and another more than 156 days.<br />(M1, D1)<br />
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