Essential Statistics Chapter 6

2 Continuous Probability Distributions chapter 5 focused upon discrete probability distributions, we now wish to focus on continuous probability distributions a normal distribution is where a continuous random variable has a distribution such that its graph is symmetric (about its mean) and is bell shaped a density curve is a graph of a continuous probability distribution, such that the area under the curve = 1, e.g. Σ P(x) = 1 every point on the curve must have a positive height value, e.g. 0 <= y <= 1

3 Continuous Probability Distributions because the total area under the density curve is = 1, there is a direct relationship between probability and area (recall length * width) uniform distribution yields a rectangle y =?

4 Continuous Probability Distributions density curve of a normal distribution has a more complicated bell shape, thus it is more difficult to calculate the probability however, there is still a correspondence between probability and area

5 Normal Continuous Distributions standard normal distribution is a normal probability that has mean of zero, μ = 0 median = mean = mode standard deviation of 1, σ = 1 area under density curve = 1 follows the empirical rule, as below

6 Normal Continuous Distributions we will find area under normal continuous distribution curves using z-scores recall z = x µ σ

7 Normal Continuous Distributions standard normal only, μ = 0, σ = 1 values left of μ have negative z-scores values right of μ have positive z-scores the area is the cumulative area from the absolute left to a vertical boundary along the x axis (the z-score)

8 Finding Area Using z-values we will use table A-2 to find z-scores for normal cont. distributions left part of table will find negative z-scores, right part will find positive z-scores three scenarios will be used 1. area up to (to the left) the z-score (less than or below) 2. area to the right of the z-score (greater than or above) 3. area between two z-scores

9 Finding Area Using z-values Scenario 1 find the area corresponding to a z-value of z = 1.26 using table A-2, we see

10 Finding Area Using z-values Scenario 2 find the area greater than a z-value of z = -0.58 using table A-2, we see

11 Finding Area Using z-values Scenario 2, cont. d we find.2810 from table, but is this what we want? since we know that the entire area under the curve is = 1, the area we want must be 1 area to the left (.281) so area shaded is 1 -.281 = 0.719

12 Finding Area Using z-values Scenario 3 find the area between z = -1.45 and z = 0.42 find the area at each of the z-values area left of z = -1.45 = 0.0735 area left of z = 0.42 = 0.6628 subtract (the larger from the smaller) to find the answer area = 0.6628 0.0735 = 0.5893

13 Finding Area Using z-values Suggestions understand what you are trying to find, and sketch a graph, indicating what area you wish to identify determine which of the three techniques you should use follow the appropriate steps for specific technique use the correct table A-2 portion (left for < 0, right for > 0)

14 Finding z-values Given Area Scenario 1 finding z-score given area to the left understand what you are trying to find, and sketch a graph, indicating what area you were given find the best approximation closest to your area, from the body of the table

15 Finding z-values Given Area Scenario 2 finding z-score given area to the right understand what you are trying to find, and sketch a graph, indicating what area you were given find the area to the left, which is 1 (area to the right) find the best approximation closest to your area, from the body of the table see special notation zα (z sub alpha) pg. 235

16 Finding z-values Given Area Scenario 3 finding z-score given area in the middle understand what you are trying to find, and sketch a graph, indicating what area you were given label two missing z values z1 and z2 find area in both tails, that is 1 (area in middle) each tail is then divided by 2 (since there are 2 of them)

17 Finding z-values Given Area Scenario 3, cont. d z1 is a simple table lookup z2 area = 1 (area in rightmost tail), look this value up in table if curve is symmetric, and z1 and z2 are equidistant from the median, z1 is negative of z2

18 Finding z-values Given Area Example 6.11: Find the z-scores that bound the middle 95% of area under standard normal curve?

19 Finding z-values Given Area Find the z-scores that bound the middle 95% of area under standard normal curve? if middle area =.95, the non-shaded white area =?

20 Finding z-values Given Area Find the z-scores that bound the middle 95% of area under standard normal curve? if middle area =.95, the non-shaded white area =.05 so each non-shaded area in each tail =.05 / 2 = o.025

21 Finding z-values Given Area Find the z-scores that bound the middle 95% of area under standard normal curve? z1: lookup area 0.025 in Table A-2, find z1 = -1.96 z2: lookup area 0.975 (.025 +.95) in table A-2, find z2 = 1.96 (or if symmetric, z2 is negative of z1)

22 Finding z-values for any Mean (6.2) in section 6.1, we found areas under curves with mean = 0 and std. dev. = 1 we will now use our z-score formula for normal distributions with any mean and std. deviations z = x µ σ z-score will indicate how many standard deviations the original value is away (above or below) from the mean

23 Finding z-values for any Mean (6.2) Example 6.15 Length of pregnancy from conception to birth is approximately normally distributed with mean μμ = 272 days and standard deviation σ = 9 days. What proportion of pregnancies last longer than 280 days?

24 Finding z-values for any Mean (6.2) find the z-score for x = 280 z = (280 272) / 9 = 0.89 look up 0.89 in table = 0.8133, this is area of un-shaded portion of graph find the shaded portion 1 0.8133 = 0.1867 so proportion = 0.1867

25 Finding Value for Given Proportion finding x given mean, std. dev. and z-value manipulating z formula, we get x = μμ + zσ see examples in text

26 Finding Value for Given Proportion Mensa is an organization whose membership is limited to people whose IQ is in the top 2% of the population. Assume that scores on an IQ test are normally distributed with mean μμ = 100 and standard deviation σ = 15. What is the minimum score needed to qualify for membership in Mensa?

27 Finding Value for Given Proportion Assume that scores on an IQ test are normally distributed with mean μμ = 100 and standard deviation σ = 15. What is the minimum score needed to qualify for membership in Mensa? if shaded area to right = 0.02, area to left must equal 1 -.02 =.98

28 Finding Value for Given Proportion Assume that scores on an IQ test are normally distributed with mean μμ = 100 and standard deviation σ = 15. What is the minimum score needed to qualify for membership in Mensa? if shaded area to right = 0.02, area to left must equal 1 -.02 =.98

29 Finding Value for Given Proportion looking up area of.98 in Table A-2, we come close at 0.9798, so z approximates to z = 2.05 the IQ score that separates the upper 2% from the lower 98% is x = μμ + zσ = 100 + (2.05)(15) = 130.75

30 Central Limit Theorem (6-3) given the following probability distribution note the skewed nature of the population histogram

31 Central Limit Theorem (6-3) notice how the histogram is reasonably well approximated as sample size grows

32 Central Limit Theorem (6-3) see Central Limit Theorem handout (or web link) if all possible random samples of size n are selected from a population with mean μ and std. deviation σ, the mean of the sample means is denoted by and standard deviation of the sample means is, referred to as the standard error of the mean

33 Central Limit Theorem (6-3) General Solution Strategy if problem is asking for an individual value, or if population is normally distributed find x, μ, σ use simple z formula if problem is asking for sample data, and original data is normal, or n > 30 find n, x bar, μ, σ use CLT z formula