Standard Deviation, Normal Distribution and Random Samples

Transcription

1 Standard Deviation, Normal Distribution and Random Samples Key knowledge The concept of sample and population and the use of random numbers as a means of selecting a simple random sample of data from a population. Statistical concepts and techniques that apply m a given practical situation for univariate data. The operating characteristics and capabilities of a CAS calculator in displaying univariatedata, calculating the value of the standard deviation and generating random numbers. You will learn to Obtain a simple random sample from a given population using a table of random numbers or an alternative random number generator (e.g. CAS calculator) Use a range of appropriate statistical techniques to analyse a practical situation for univariate data and to draw valid conclusions from this analysis. Use appropriate functions of a CAS calculator to display univariate data, to calculate the standard deviation and to generate random numbers Tasks relate to Outcomes 1, 2 and 3.

2 SUMMARISING DATA A statistic is any number that can be computed from data. Certain special statistics are called summary statistics, because they numerically summarise certain features of the data set. Summary statistics are generally either measures of centre (which you have already learnt) or measures of spread. There are many different examples for each of these measures, and there are situations when one of the measures is more appropriate than another. MEASURES OF SPREAD A measure of spread is calculated in order to judge the variability of a data set. That is, are most of the values clustered together, or are they rather spread out? You have already looked at the range and interquartile range as measures of spread in data. The range measured the difference between the highest and lowest values. The interquartile range measured the spread of the data about the median. It is the difference between Q 1 and Q 3. IQR = Q3 Q 1 STANDARD DEVIATION The standard deviation is another measure of spread of a data set and has wider applications in statistics. The standard deviation(s) is a summary statistics that describes the spread of the data around the mean (x). The mean (x) is also a summary statistics that can be used to locate the centre of a symmetric distribution. Formula for finding the Standard Deviation for Ungrouped Data s = (x x) 2 n 1 x is the mean x represents the data values in the sample n is the number of values in the sample The standard deviation is the square root of the average of the squared deviations from the mean.

3 Finding Standard Deviation for Ungrouped Data Example 1 Work out the standard deviation correct to 3 decimal places of the heights (in cm) of twelve people. 152, 175, 163, 181, 133, 145, 159, 162, 165,173, 176, 154 Step 1: Calculate the mean. 𝑥 = 𝑥 1938 = = 𝑤ℎ𝑒𝑟𝑒 𝑛 12 𝑥 = 𝑠𝑢𝑚 𝑜𝑓 𝑡ℎ𝑒 𝑑𝑎𝑡𝑎 𝑣𝑎𝑙𝑢𝑒𝑠 𝑖. 𝑒 Step 2: Calculate the standard deviation. i. ii. iii. Calculate the deviations from the mean by subtracting the mean from each x value i.e. find (𝑥 𝑥) Square the deviations i.e. find (𝑥 𝑥)2 Substitute into the formula below Height 𝒙 𝑠= 𝑥 𝑥! 𝑛 1 𝑠= 𝑠 = 𝑠 = (3𝑑𝑝) The standard deviation is 𝑥 = 1938 (𝒙 𝒙) (𝒙 𝒙)𝟐 (𝑥 𝑥) = 0 (𝑥 𝑥)! = 2157 The above method is tedious and time consuming. It is easy to make careless errors when there are many calculation involved. Thus, you should just use the CAS calculator to find the standard deviation of any given ungrouped set of data. Standard deviation can be approximated with the rule: 𝑠! = 𝑟𝑎𝑛𝑔𝑒 4 Round result to nearest whole number. It is used to gauge reasonableness.

4 The calculator gives the following information: x x x 2 sx σx n minx Q 1 Med Q 3 maxx Mode ModeN ModeF = mean = the sum of the data = the sum of the squares of the data = the population standard deviation = the sample standard deviation = the sample size (number of data values) = the smallest data entry (minimum) = data at the first quartile = data at the median (second quartile) = data at the third quartile = the largest data entry (maximum) = mode (most frequent data value) = number of data mode items = data mode frequency Exercise 3B Remember: x = the mean of a sample and µ = the mean of a population s x = the standard deviation of a sample and σ x = the standard deviation of a population s 2 = the variance of a sample and σ 2 = the variance of a population s = variance of sample and σ = variance of a population

5 Normal Distribution: % rule The histogram below represents the heights of a large group of adult students, We can see that the data is approximately symmetric. On top of the histogram a smooth curve has been drawn. The curve is bell-shaped. Sets of data which have this shape are called normal distributions. Some examples of sets of data which are normally distributed are: Height Weight Blood pressure readings IQ scores If a set of data is normally distributed (bell-shaped), then there is important information that is known about the data. mean x 1. 68% of the data lies within one standard deviation of the mean x s to x + s 2. 95% of the data lies within two standard deviations of the mean x 2s to x + 2s % of the data lies within three standard deviations of the mean x 3s to x + 3s These intervals are also known CONFIDENCE INTERVALS

6 Example 1 A survey of weights of office workers formed a bell-shaped frequency curve. The mean weight is 61 kg with a standard deviation of 7.4 kg. Find the confidence intervals in which you would find: (a) 68% of weights? (b) 95% of weights? (c) 99.7% of weights? Solution Mean: x = 61 Standard deviation: s = 7.4 (a) 68% of the weights lie in the interval x s to x + s x s = = 53.6 x + s = = 68.4 This means that about 68% of weights he between 53.6 kg and 68.4 kg (b) 95% of the weights lie in the interval x 2s to x + 2s x 2s = = 46.2 x + 2s = = 75.8 This means that about 95% of weights he between 46.2 kg and 75.8 kg. (b) 99.7% of the weights lie in the interval x 3s to x + 3s x 3s = = 38.8 x + 3s = = 83.2 This means that about 99.7% of weights he between 38.8 kg and 83.2 kg 68% 95% 99.7%

7 Other Percentages If 68% of the values lie within one standard deviation of the mean then 32% of the values lie outside this range. This means that 16% lie in each of the tails. 16% of data 16% of data 68% x! - s x! x! + s If 95% of the values lie within two standard deviations of the mean then 5% of the values lie outside this range This means that 2.5% lie in each of the tails. 2.5% of data 95% 2.5% of data x! - 2s x! x! + 2s If 99.7% of the values lie within three standard deviations of the mean then 0.3% of the values lie outside this range. This means that 0.15% lie in each of the tails. 99.7% 0.15% of data 16% of data x! - 3s x! x! + 3s Because the normal distribution is symmetrical then 50% of the values lie above the mean and 50% below the mean. 50% of data 50% of data x! (mean) and median The mean and the median coincide in a perfectly normal distribution!

8 Example 2 The scores of a test are normally distributed with a mean of 28 and a standard deviation of 4. (a) What percentage of students has a score between 24 and 32? (b) What percentage of students has a score between 20 and 36? (c) What percentage of students has a score between 16 and 40? (d) What percentage of students has a score less than 20? (e) What percentage of students has a score greater than 20? (f) What percentage of students has a score less than the average score? (g) What percentage of students has a score greater than 32? (h) If 6000 students sit the test how many would receive a score greater than 32? Remember: Always draw a bell curve first showing mean and confidence intervals. 68% 95% 99.7% Use the % rule to calculate the following: (a) What percentage of students has a score between 24 and 32? These scores are within one standard deviation of the mean. Therefore 68% of students have a score between 24 and 32 (b) What percentage of students has a score between 20 and 36? These scores are within two standard deviations of the mean. Therefore 95% of students have a score between 20 and 36. (c) What percentage of students has a score between 16 and 40? These scores are within three standard deviations of the mean. Therefore 99.7% of students have a score between 16 and 40.

9 (d) What percentage of students has a score less than 20? Scores less than 20 are more than two standard deviations below the mean. Therefore 2.5% of scores are less than 20. (e) What percentage of students has a score greater than 20? Use your result from part (d). If 2.5% of scores lie below 20 then 100% - 2.5% = 97.5% of scores are greater than 20. (f) What percentage of students has a score less than the average score? 28 is the mean or average score. Therefore 50% of the scores lie below 28, the average score. (g) What percentage of students has a score greater than 32? Scores greater than 32 are more than one standard deviation above the mean. 16% of scores are greater than 32. (h) If 6000 students sit the test how many would receive a score greater than 32? Find 16% of % of 6000 = 960 You would expect about 960 students to receive a score greater than 32 lowest score mean highest score 16% 34% 34% 68% x s x + s 16% 2.5% 47.5% 47.5% 95% 13.5% 13.5% x 2s x + 2s 2.5% 0.15% 49.85% 49.85% 99.7% 2.35% 2.35% x 3s x + 3s 0.15% 0% 0.15% 2.5% 16% 50% 84% 97.5% 99.85% 100% z- scores Note: 95% rule can be used on any distribution i.e. skewed data Exercises 3C

10 STANDARD SCORES (z-scores) The significance of a particular data value can be looked at in terms of the number of standard deviations it lies from the mean. A normally distributed set of data can be transformed into a new set of data This technique is called standardisation. The new data values are called standard scores or z-scores A standard score (z score) tells us how many standard deviations each data value lies from the mean. For example, take a normal distribution which has a mean of 80 and a standard deviation of 5 𝑥 𝑠 = 80 5 = 𝟕𝟓 𝑥 𝑠2 = = 𝟕𝟎 𝑥 3𝑠 = = 𝟔𝟓 𝑥 + 𝑠 = = 𝟖𝟓 𝑥 + 2𝑠 = = 𝟗𝟎 𝑥 + 3𝑠 = = 𝟗𝟓 68% of values lie between 75 and 85 95% of values lie between 70 and % of values lie between 65 and 95 FORMULA FOR CALCULATING STANDARD SCORES (Z-SCORES) 𝒅𝒂𝒕𝒂 𝒗𝒂𝒍𝒖𝒆 𝒎𝒆𝒂𝒏 𝒔𝒕𝒂𝒏𝒅𝒂𝒓𝒅 𝒅𝒆𝒗𝒊𝒂𝒕𝒊𝒐𝒏 𝒙 𝒙 𝒛 = 𝒔 𝒔𝒕𝒂𝒏𝒅𝒂𝒓𝒅 𝒔𝒄𝒐𝒓𝒆 (𝒛 𝒔𝒄𝒐𝒓𝒆) = 𝒙 𝒊𝒔 𝒕𝒉𝒆 𝒎𝒆𝒂𝒏 𝒙 𝒓𝒆𝒑𝒓𝒆𝒔𝒆𝒏𝒕𝒔 𝒕𝒉𝒆 𝒅𝒂𝒕𝒂 𝒗𝒂𝒍𝒖𝒆 𝒔 𝒊𝒔 𝒕𝒉𝒆 𝒔𝒕𝒂𝒏𝒅𝒂𝒓𝒅 𝒅𝒆𝒗𝒊𝒂𝒕𝒊𝒐𝒏 To find a z-score subtract the mean from the data value and divide by the standard deviation.

11 From our example find the z-score for 75. x = 75 x = 80 s = 5 x x z = s z = 5 z = 5 5 z = 1 75 is equivalent to a z-score of -1. i.e. it is one standard deviation below the mean which we know from the above calculations. The standard scores for all the data-values are shown in the table below. Original value z-score From the table we can see: If a data value is one standard deviation ABOVE the mean then it has a z-score of 1 If a data value is one standard deviation BELOW the mean then it has a z-score of -1 If a data value is two standard deviations ABOVE the mean then it has a z-score of 2 If a data value is two standard deviations BELOW the mean then it has a z-score of -2 If a data value is three standard deviations ABOVE the mean then it has a z-score of 3 If a data value is three standard deviations BELOW the mean then it has a z-score of -3 What is the z-score for the mean? x = 80 x = 80 s = 5 x x z = s z = 5 z = 0 The standardized score for the mean is zero i.e. z-score for mean = 0

12 Look at the two graphs and compare the x data values with the corresponding standardised z-scores. The interval x ± 2s is shown When x = 65, z = -3 When x = 70, z = -2 When x = 75, z = -1 When x = 80, z = 0 When x = 85, z = 1 When x = 90, z = 2 When x = 95, z = The same percentages apply to the standardised scores 68% of values have z-scores between -1 and 1 95% of values have z-scores between -2 and % of values have z-scores between -3 and 3 A positive z-score indicates the data value lies above the mean A negative z-score indicates the data value lies below the mean A zero z-score indicates the data value is equal to the mean Example 1 Find the standard score or z-scores for each of the following x values for a distribution with mean of 63 and a standard deviation of 2.9. Round off your answers to 2 decimal places. (a) x = 59 x x z = = s = A z-score of tells us that the data value 59 is 1.38 standard deviations below (since it is negative) the mean. It is in the bottom 16% of values. (b) x = 72.5 x x z = = s = A z-score of 3.28 tells us that the data value is 3.28 standard deviations above (since it is positive) the mean. It is in the top 0.15% of values.

13 Example 2 (a) A data value from a normal distribution has a z-score of If the distribution has a mean of 40 and a standard deviation of 2.3, find the value of the data value. 𝑥 𝑥 𝑠 𝑥 = 2.3 𝑧= = 𝑥 = 𝑥 40 𝑥 = 𝑥 = The data value is (b) Find the mean of a normal distribution if a data value of 60 has a z-score of 0.8, and the standard deviation of the distribution is 7.4. 𝑧= 0.8 = 𝑥 𝑥 𝑠!"!!!.! = 60 𝑥 Note: NumSolve can be used to transpose equations on your Classpad 5.92 = 60 𝑥 𝑥 = 𝑥 = The mean of the distribution is (c) Find the standard deviation of a normal distribution if a data value of 156 has a z-score of -2.3, and the mean of the distribution is 182. Answer to three decimal places. 𝑥 𝑥 𝑠 = 𝑠 𝑧= 2.3𝑠 = 26 𝑠= 𝑠 = The standard deviation of the distribution is

14 Comparing data from different distributions Standard scores are useful when comparing scores from different sets of data. Example 3 Joan sat for two English tests during the year. Her results are shown below. In which test did she perform better? Test Joan s mark Class mean Standard deviation Use the information to standardize her scores. Test 1 Test 2 x x x x z = = = 3.17 (2 dp) z = = = 3.25 (2 dp) s 6 s 4 Her result for Test 1 was 3.17 standard deviations above the mean. Her result for Test 2 was 3.25 standard deviations above the mean. Hence Joan performed better in Test 2. Both marks were in the top 0.15% of the class.

15 Example 4 A group of 50 students run a 100 m and a m race. One student has times of seconds for the 100 m and 43 minutes 29 seconds (2609 seconds) for the m. Both these times are better than the average times for the distances but comparatively which performance is better? 100 metres metres x x s Calculate the z-scores for each time: x x z = s = = (3 dp) x x z = s = 953 = (3 dp) These z-scores imply that the student s time for the m is comparatively better. Being more than 1 standard deviation below the mean, this time is in the top 16% of times for the m. (The lower the time he better the result.) The student s time for the 100 m is within the middle 68% of times, that is, ± 1 standard deviation from the mean. Exercise 3D

16 RANDOM SAMPLES If data is to be collected for a survey, it is important that the data reflects the whole population. For example, a survey is to be conducted on all Year 9 students The survey is looking at the number of books each student reads in a year. It is impractical to survey all Year 9 students so a sample is taken from the population (all Year 9 students) It must be collected from a representative sample so that the statistics calculated can be applied to the whole population. When choosing a simple random sample ensure that: The sample must be representative of the population The sample must be random so that each member of the population must have an equal chance of being chosen The selection of one person from the population should not affect the selection of another person ANALYSING SIMPLE RANDOM SAMPLES Once we have a simple random sample we can calculate statistics. Example 1 Remember: Always check if finding sample or population. DATA SET ONE on the next page shows the number of days that thunderstorms occur in a year for 81 American cities We will use this as our population. (a) (i) Generate 10 random numbers in the range 1 to 81 (inclusive). Use these numbers to obtain your sample of 10 cities. (ii) Use the calculator again to generate a second sample of ten. (b) Calculate the following summary statistics for each sample: (i) mean (iii) mode (ii) median (iv) range (v) Q 1 (vi) Q 3 (vii) IQR (viii) standard deviation (c) Compare the results of your first and second samples. Sample solution

17

18 COMPARISONS BETWEEN THE TWO BOXPLOTS Using the box plots, we can compare the two samples. It is seen that the statistics for each sample are quite different. The very large range and standard deviation for the first sample in particular, suggests a lot of variability in the data. To make results more reliable, a larger sample may need to be taken. Similarities Neither distribution has outliers The medians of both distributions are offcentre and are close to the upper quartile indicating some degree of negative skewedness. Differences The median of the first sample is much greater than that of the second sample. The range and interquartile range for sample 1 are both much higher than for sample 2. Thus the spread of sample 1 is much greater than that of sample 2. Exercise 3E