Presentation on theme: "Lecture 7 THE NORMAL AND STANDARD NORMAL DISTRIBUTIONS"— Presentation transcript:
1Lecture 7 THE NORMAL AND STANDARD NORMAL DISTRIBUTIONS
2Populations and samples When we gather data, the POPULATION is the reference set containing ALL POSSIBLE OBSERVATIONS (ALL scores, ALL reaction times, ALL IQs).Our own data are usually a selection or SAMPLE from the population.In statistics, our data are assumed to be samples from known THEORETICAL populations.
4A sample from a population This is a picture of the distribution of 4000 IQs.The histogram is symmetrical and bell-shaped.That’s because we have sampled from a NORMAL DISTRIBUTION.The normal distribution is the most important theoretical population in statistics.
5Large samplesFrom the Laws of Large Numbers, we can expect the values of sample statistics to be close to those of the corresponding population parameters.The mean of this large sample is 99.9 and the SD is These values are quite close to the population mean of 100 and the population SD of 15.
6It makes sense to say that, if IQ is normally distributed with a mean of 100 and an SD of 15, we have sampled 4000 values from a normal distribution.
7‘Population’ means ‘distribution’ In these lectures, I shall use the terms ‘population’ and ‘distribution’ interchangeably.
8Statistics versus parameters STATISTICS are characteristics of SAMPLES; PARAMETERS are characteristics of POPULATIONS.A normal population has two parameters:the mean;the standard deviation.The IQ population is (approximately) a normal distribution with a mean of 100 and an SD of 15.
9A notational convention Letters from the Roman alphabet, such as M and s (for the mean and standard deviation, respectively), are used to denote the values of the STATISTICS of samples.Greek letters, (μ, σ) are used to denote the values of the corresponding population characteristics or PARAMETERS.
10The IQ exampleIn this particular sample of 4000 IQ’s, M = 99.9 and s =In the population, μ = 100 and σ = 15.
11The caffeine experiment In the caffeine experiment, we are sampling from TWO populations:the population of scores under the Placebo condition with mean μ1 and standard deviation σ1;the population of scores under the Caffeine condition with mean μ2 and standard deviation σ2.
12Specifying a normal distribution Suppose that a variable X has a normal distribution with mean μ and standard deviation σ.We write this as shown.
13There are many normal distributions There are an infinite number of normal distributions, each specified by particular values for μ and σ.The IQ is approximately distributed as N(100, 15).The heights of men are approximately distributed as N(69, 2.6).
14IQs of at least 130Suppose IQ has a normal distribution, with a mean of 100 and a standard deviation of 15.What proportion of people have IQ’s of at least 130?
15At least 130? …If a variable is normally distributed, 95% of values lie within 1.96 standard deviations (2 approx.) on EITHER side of the mean.So only 2 ½ % (0.025) of values lie beyond 2 SD’s above the mean.0.95 (95%)2 ½ % = .0252 ½ % = .025meanmean – 1.96×SDmean +1.96×SD
16At least 130? … An IQ of 130 is 2 standard deviations above the mean. So only 2 ½ % (0.025) of IQ’s lie beyond 130.0.95 (95%)2 ½ % = .0252 ½ % = .025meanmean – 1.96×SDmean +1.96×SD
17ProbabilityA PROBABILITY is a measure of likelihood ranging from 0 (an impossibility) to 1 (a certainty).In the POPULATION, relative frequencies are probabilities.
18Relative frequency as an area Write a little into this boxRelative frequency of heights between 65 inches and 70 inches.
19Probability In the POPULATION, relative frequencies are PROBABILITIES. The area under the normal curve of height between 65 inches and 70 inches is the PROBABILITY of a height in that range.
20Probability as an area Write a little into this box Probability of a height between 65 inches and 70 inches.
21IQ and probabilityIf IQ is indeed normally distributed with a mean of 100 and an SD of 15, 2.5% of values in the population are greater than 130.The PROBABILITY of an IQ of at least 130 isProbability of an IQ greater than 130= 0.0250.950.025
22NotationIntelligence is assumed to have a CONTINUOUS DISTRIBUTION: there are an infinite number of values between any two points.So the probability of any one value is zero.Consequently Pr(IQ ≥ 130) = Pr(IQ > 130) and Pr(IQ ≤ 130) = Pr (IQ < 130).
23Probability densityAssociated with each value x of IQ is a PROBABILITY DENSITY, which can be thought of as the probability of a value IN THE NEIGHBOURHOOD of x.The height of the normal curve above the value x is the probability density of x.
24Probability distribution The normal distribution is an example of a PROBABILITY DISTRIBUTION.It is so-called because we can use it to obtain the probability of values of the variable within a specified range.There are several important probability distributions in statistics, and they are all used for this purpose.
25The standard normal variable z To find out how many standard deviations an IQ of 130 is above the mean, we have to SUBTRACT the mean and DIVIDE by the value of the standard deviation, i.e., by 15.If X is the original variable (X is IQ in this example), we have transformed X into another variable z, which is known as the STANDARD NORMAL VARIABLE.
26The standard normal variable z If X is a normal variable, that is,X~N(μ,σ),z will also be normally distributed.z is known as the STANDARD NORMAL VARIABLE.
27StandardisationStrictly speaking, z is defined in relation to the theoretical population mean μ.However, any set of scores X can be STANDARDISED by subtracting the sample mean from each score and dividing by the sample standard deviation.We shall investigate the effects of standardising the 4000 IQ scores in our large sample.
32Distribution of zThe distribution of z is also normal, but it is centred on zero and has a standard deviation of 1.
33Scientific notation -1.4016E-4 means ×10-4 = , which is zero, within rounding error.
34The statistics of zJust use the Descriptives procedure to find the mean and standard deviation of z.The mean is 0.The standard deviation is 1.
35Effects of standardisation Standardising a set of scores (or a population of scores) has two effects:The mean becomes zero;The standard deviation becomes 1.
36The standard normal distribution In the notation I introduced earlier, we can represent the standard normal distribution as follows.
37Distribution of zStandardising a set of scores does NOT make them normally distributed.If there’s a tail to the right (+ve skew) before transforming X to z, there will be one after the transformation.Nevertheless, whatever the shape of the original distribution, the mean standardised scores will be zero and the standard deviation will be 1.
38Deviations sum to zeroZero deviations-ve deviations+ve deviationsThe mean is the centre of gravity, or balance point. The deviations are the distances of the points from the balance point. They must sum to zero: the positives and negatives must cancel each other out.
39The mean of z The numerator of z, (X – mean), is a DEVIATION SCORE. Since deviations about the mean sum to zero, the mean of the distribution of z is also zero.So the bell-shaped STANDARD NORMAL DISTRIBUTION is centered on zero.
41Using z Probability that X (IQ) lies between 70 and 130 AND ALSOProbability that z lies between andProbability that X (IQ) is at least 130AND ALSOProbability that z is at least +1.960.95X (IQ) = SDz
42Referring questions from X to z What is the probability of an IQ of at least 130?This is to ask about the probability that X is at least 130, where X ~N(100, 15).Transform X to z: z = (130 – 100)/15 = 2.We know that the probability of z greater than 2 (1.96) = .025.
43Between 100 and 130? Convert these values to values of z. If X = 100, z = 0.If X = 30, z = 2.Pr(z between 0 and 2) =0.95 (95%)2 ½ % = .0252 ½ % = .025μX μ – 1.96×SDμ +1.96×SDz
44Finding the probability of a range of values of X In the problems we have considered, the value of z has always been around +2 (about 1.96), so that we can find the probability from memory.Suppose z = 1, 0.5, or any value other than 1.96?Just standardise the value of X by converting it to z: z = (X – mean)/SD.The are available tables in standard statistics textbooks which give probabilities of any specified range of z. You can also use SPSS to find such probabilities.
45The standard normal distribution There are countless normal distributions.But there is only ONE standard normal distribution, to which any of the others can be transformed by z = (X – mean)/SD.So only the probabilities of ranges of values of z need to be tabled.It would not be feasible to table the probabilities for ALL possible normal distributions.
46To sum up …If we know the DISTRIBUTION of some variable, we can assign a probability of obtaining a value within a specified range.We can visualise the probability of such a value as the area under the curve of the distribution.If the distribution is normal, we can translate probability questions in the original units of measurement into questions about ranges of z, which, provided X is normally distributed, has the STANDARD NORMAL DISTRIBUTION.
47PercentilesA PERCENTILE is the VALUE or SCORE below which a specified percentage or proportion of the distribution lies.The 30th percentile is the value below which 30% of the distribution lies.The 70th percentile is the value below which 70% of scores lie.
48The 30th and 70th percentiles (0.70)0.3030th percentile0.70(0.30)70th percentile
49Cumulative probability The CUMULATIVE PROBABILITY of a particular value is the probability of a value LESS THAN OR EQUAL TO the value.The cumulative probability of a value at the 30th percentile is The cumulative probability of a value at the 70th percentile is 0.70.Cumulative probability of a score at the 30th percentile= 0.3030thCumulative probability of a score at the 70th percentile= 0.70.70th
50Using cumulative probabilities Given that height is normally distributed, with a mean of 69 inches and an SD of 2.6 inches, what is the probability of a man having a height between 65 and 70 inches?
51CumProb (65)65←CumProb (70)→70Pr of height between 65 & 70
59SPSS exercise 1 Open the SPSS data file containing 4000 IQ’s. Use the Compute procedure (in the Transform menu) to standardise the scores.Use Descriptives to obtain the mean and standard deviation of z.
60SPSS exercise 2Assuming that height is normally distributed with a mean of 69 inches and an SD of 2.6 inches, what is the probability of a man having a height between 74.2 inches and 76.8 inches?Solve by using the CDF to find the cumulative probabilities directly and subtracting.Transform the heights to z and compare the cumulative probabilities you obtain with those you obtained using the first approach.