Upcoming SlideShare
×

# Normal lecture

1,135 views

Published on

Normal Lecture

Published in: Education
0 Likes
Statistics
Notes
• Full Name
Comment goes here.

Are you sure you want to Yes No
• Be the first to comment

• Be the first to like this

Views
Total views
1,135
On SlideShare
0
From Embeds
0
Number of Embeds
6
Actions
Shares
0
12
0
Likes
0
Embeds 0
No embeds

No notes for slide

### Normal lecture

1. 1. http://www.youtube.com/watch?v=NYd6wz YkQIM&feature=relmfuThis is the most important and mostwidely used probability distribution 1
2. 2. The Normal Distribution• Properties – Bell shaped 1 – Area under curve equals 1 – Symmetric around the mean μ – Mean = median = Mode – Two tails approach the horizontal axis – never touch axis – Empirical rule applies – Two parameters – μ and σ 2
3. 3. How does the standard deviation affect the shape of the distribution s=2 s =3 s =4 m = 11How does the mean affect the location of the distribution s=2 m = 10 m = 11 m = 12 3
4. 4. The Normal DistributionMathematical model expressed as: 1  xm  2 1   2 s f ( x)  e ,   x    2swhere   3.14159 and e  2.71828- notation N(μ ; σ2) 4
5. 5. STANDARDISING THE RANDOM VARIABLE• Seen how different means and std dev’s generate different normal distributions• This means a very large number of probability tables would be needed to provide all possible probabilities• We therefore standardise the random variable x so that only one set of tables is needed 5
6. 6. STANDARDISING THE RANDOM VARIABLE• A normal random variable x can be converted to a standard normal variable (denoted Z) by using the following standardisation formula:- xmz , x any value of the random variable X s 6
7. 7. The Standard Normal Distribution• Different values of μ and σ generate different normal distributions• The random variable X can be standardised – mean = μ = 0 – standard deviation = σ = 1 xmz , x any value of the random variable X s 7
8. 8. The Standard Normal Distribution μ=0 s=1 z -3 -2 -1 0 1 2 3 z-values on the horizontal axis • distance between the mean and the point represented by z in terms of standard deviation 8
9. 9. Finding Normal Probabilities• Example – Marks for a semester test is normally distributed, with a mean of 60 and a standard deviation of 8 – X ~ N(60 , 82) – If we need to determine the probability that the mark will x 50 60 65 be between 50 and 65, we need to determine the size of the shaded area – Before calculating the probabilities the x-values need to be transformed to z-values 9
10. 10. P(-∞ < Z < z)Standard normal probabilities have beencalculated and are provided in a tableThe tabulated probabilitiescorrespond to the area between 0 zZ = -∞ and some z > 0 z 0.00 0.01 → 0.05 0.06 → 0.09 0.0 0.5000 0.5040 0.5199 0.5239 0.5359 0.1 0.5398 0.5438 0.5596 0.5636 0.5753 ↓ 1.0 0.8413 0.8438 0.8531 0.8554 0.8621 1.1 0.8643 0.8665 0.8749 0.8770 0.8830 1.2 0.8849 0.8869 0.8944 0.8962 0.9015 10 ↓
11. 11. • Example continue – If X denotes the test mark, we seek the probability – P(50 < X < 65) – Transform the X to the standard normal variable ZEvery normal variable X m Therefore, once with some m and s, Z probabilities for Z are can be transformed s calculated, probabilities into this Z E(Z) V(Z) of any normal variable 11 μ=0 σ2 = 1 can be found
12. 12. • Example continue Mean = μ = 60 minutes Standard deviation = σ = 8 minutes X m 50 -- 60 < Z < 65 -- 60 ) X mP(50 < X < 65) = P( 8s 8s = P(-1.25 < Z < 0.63) X-m Z= sTo complete the calculation we need to computethe probability under the standard normal distribution12
13. 13. • How to use the z-table to calculate probabilities ExampleDetermine the following probability: P(Z > 1.05) = ? 1 - P(Z < 1.05) = P(Z > 1.05) 0 1.05 P(Z > 1.05) = 1 – 0.8531 = 0.1469 13
14. 14. • How to use the z-table to calculate probabilitiesExampleDetermine the following probability: P(-2.12 < Z < 1.32) = ? 0.9066 P(-∞ < Z < 1.32) = 0.9066 0,5 1.32 P(-2.12 < Z < +∞) = 0.9830 0.9830 0,5 -2.12 + -2.12 0 1.32 P(-2.12 < Z < 1.32) = (0.9066 + 0.9830) - 1 = 0.8896 14
15. 15. • Example – Determine the following probabilities: P(0.73 < Z < 1.40) = ? P(-∞ < Z < 1.40) = 0.9192 0.9192 1.40 P(-∞ < Z < 0.73) = 0.7673 0.7673 0 0.73 1.40 0.73P(0.73 < Z < 1.40) = 0.9192 – 0.7673 = 0.1519 15
16. 16. The symmetry of the normal distributionmakes it possible to calculate probabilitiesfor negative values of Z using the table asfollows: P(-z < Z < +∞) = P(-∞ < Z < z) 16
17. 17. • Example student marks - continued P(50 < X < 65) = P(-1.25 < Z < 0.63) 0.7357 0.8944 = 0.8944 + 0.7357 – 1 0.8944 0.8944 0.8944 0.8944 0.8944 = 0.6301 z = 0.63 In this example z = -1.25 , because of symmetry read 1.25 z 0.00 0.01 → 0.05 0.06 → 0.09 0.0 0.5000 0.5040 0.5199 0.5239 0.5359 0.1 0.5398 0.5438 0.5596 0.5636 0.5753 ↓ 1.0 0.8413 0.8438 0.8531 0.8554 0.8621 1.1 0.8643 0.8665 0.8749 0.8770 0.8830 1.2 0.8849 0.8869 0.8944 0.8962 0.9015 17