Statistics 2

Find the standard deviation to the nearest tenth for the following data set: {10, 10.5, 10.3, 11, 15, 20 , 16.7}

1. Mean: 13.36 2. Data points - mean: -3.36, -2.86, -3.06, -2.36, 1.64, 6.64, 3.34 3. Squared: 11.29, 8.18, 9.36, 5.57, 2.69, 44.09, 11.16 4. Variance: 13.19 5. Square root: 3.6 = standard deviation

Calculate the standard deviation for the following numbers to the nearest tenth. {1, 3, 4, 2, 2}

1. Mean: 2.4 2. Data points- mean: 1.4, 0.6, 1.6, -0.4, -0.4 3. Squared: 1.96, 0.36, 2.56, 0.16, 0.16 4. Variance: 1.04 5. Square root: 1.0= standard deviation

Find the standard deviation to the nearest hundredth for the following data set: {3, 3, 4, 4, 5}

1. Mean: 3.8 2. Data points - mean: -0.8, -0.8, 0.2, 0.2, 1.2 3. Squared: 0.64, 0.64, 0.04, 0.04, 1.44 4. Variance: 0.56 5. Square root: 0.75 = standard deviation

Calculate the standard deviation of the following numbers to the nearest hundredth. {4, 6, 7, 8, 5, 10)

1. Mean: 6.67 2. Data points- mean: 2.67, -0.67, 1.33, 0.33, -1.67, 3.33 3. Squared: 7.13, 0.45, 1.77, 0.11, 2.79, 11.09 4. Variance: 3.89 5. Square root: 1.97 = standard deviation

Find the mean of the following data set: {100,112,120,124,131,131,108}

100 + 112 + 120 + 124 + 131 + 131 + 108 = 826/7 = 118

A group of 5 cats has a mean whisker length of 12 in. Find the whisker length of a 6th cat that would drop the mean to 11 in.

12 * 5 cats = 60 (60 + x) / 6 = 11 66 = 60 + x x = 6 The cat's whiskers would need to be 6 inches to bring the mean whisker length to 11 inches

Calculate the mean of this data set to the nearest whole number: {99, 98, 97, 76, 85, 84, 80, 95, 74, 79, 70}

99 + 98 + 97 + 76 + 85 + 84 + 80 + 95 + 74 + 79 + 70 = 937/11 = 85

bimodal distribution

A distribution (of opinions) that shows two responses being chosen about as frequently as each other.

Spread in data

A measure of how far from the middle of a data set the individual values are (Ex: range, interquartile range, variance and standard deviation)

Measures of Central tendency

A measure used to describe the central value set of data; mean, median, and mode

Simple random sample

A method of sampling in which every member has an equal chance (or probability) of being chosen

Find the mode of the following data set: 10, 14, 15,12,10, 14, 19, 12, 14

Arranged in increasing order: 10, 10, 12, 12, 14, 14, 14, 15, 19 Mode = most frequent number = 14

Standard Error

How close a sample mean is to the to the data expected from the population. To calculate: take the standard deviation of the sample and divide by the square root of sample size

Find the mode of the following data set: {1, 2, 3, 4, 1, 2, 3, 1, 5, 7}

In increasing order: 1, 1, 1, 2, 2, 3, 3, 4, 5, 7 Mode = most frequent number = 1

Find the median of the following data set: {100, 112, 120, 124, 131, 131, 108}

In increasing order: 100, 108, 112, 120, 124, 131, 131 Median = number in middle = 120

Find the mean and median for the following data set: { 180, 185, 160, 175, 210, 201, 205}

In increasing order: 160, 175, 180, 185, 205, 210, 201 Mean: 160 + 175 + 180 + 185 + 205 + 210 + 210 = 1325/7 = 189.3 Median: 185

Find the mode of the height

In increasing order: 68, 69, 70, 71, 71, 72, 72, 72, 73, 74 Mode = most frequent # = 72

Find the interquartile range for the following data set: {70, 68, 84, 85, 82, 81, 90, 92, 94, 72}

In increasing order: 68, 70, 72, 81, 82, 84, 85, 90, 92, 94 Median (quartile 2): 83 Median of 1st half (quartile 1): 72 Median pf 2nd half (quartile 3): 90 90 - 72 = 18

Find the mean and median and determine which value is the better measure of central tendency for the following data set: {20, 22, 24, 25, 26, 27, 60}

Mean: 20 + 22 + 24 + 25 + 26 + 27 + 60 = 204/7 = 29 Median: 25 The median is more appropriate because the outlier (60) is skewing the mean to the right.

Using mean vs. median

Mean: useful for finding the central tendency of data that are close together Median: useful when data include outliers to prevent skewing the analysis.

Determine if the following is a simple random sample: 10 students were chosen from a chemistry classroom to serve as a sample of the school.

Not a simple random sample: not every student in the school had an equal chance of being selected.

Calculate the median of the following set of data: {12, 14, 15, 22, 13, 45, 8}

Numbers ordered from smallest to largest: 8, 12, 13, 14, 15, 22, 45 Median = number in the middle = 14

Find the range of the following data set: {100, 112, 120, 124, 131, 131, 108}

Range = largest value - smallest value 131-100 = 31

Sample Mean

The average of the data found from the sample. To calculate: add all data points together, then divide by the total number of data points.

Interquartile range (IQR)

The difference between the value of the 1st quartile and the value of the 3rd quartile of a data set.

Mode

The value that occurs most frequently in a given data set.

Quartile

The values that separate a data set into 4 groups 2nd quartile: the median of the data 1st quartile: the median of the first half of the data 3rd quartile: the median of the second half of the data

Finding a student's percentile if they are in 4th place in a class of 20 people

p= (k + 0.5r)/ n k= 16; r = 1; n = 20 p = (16 + 0.5*1)/20 = 0.825 = 83rd percentile

The Law of Large Numbers

states that larger sample sizes lead to the sample mean being closer to the mean of the population ( a more accurate statistical representation of the population)

Mean

the arithmetic average of a distribution, obtained by adding the scores and then dividing by the number of scores

Median

the middle score in a distribution; half the scores are above it and half are below it