Transcript Chp 5
Statistics Chapter 5 Describing Distributions Numerically By: William Pham , Dayla Larocque 10 values to find • Median: the exact middle value from the data • Range: maximum – minimum, be wary of outliers as this will skew the data • Interquartile Range: found by finding the lower and upper quartiles then subtracting the larger from the smaller. • Lower quartile: the lower 25% of the given data • Upper quartile: the upper 25% of the given data • The maximum: the highest value of the given data • The minimum: the lowest value of the given data • Mean: 10 Values to find cont. • Variance: sum of squared deviations from the mean divided by the count minus one ie: s 2 y y 2 n 1 2 • Simply standard deviation = (sum of difference 2 between mean and value) divided by the number of terms minus 1 10 Values to find cont. • Standard deviation: square root of the variance ie: s y y 2 n 1 • Or the square root of the sum of values of the mean minus given squared divided by the number of values minus one Graphs to encounter • • • • • Graphs encountered include: boxplots bar graphs Histograms Stem and leaf plots Line graphs What to do with the data • Based on given data from tables you can reform into a boxplot • To do so put all data into a list from least to greatest value. • Find the min and max median and mean • Then find the IQR by finding the lower and upper quartile then subtracting lower from upper • Find then variance given the other values • After finding variance calculate standard deviation What to do with the data cont • Given all of the values calculate variance • After finding variance you can find the standard deviation • Finally after finding all relevant values examine and report any details and report on the spread and as well as outliers • See if the graph is skewed, or symmetric Possible issues • Technology used may be off due to outliers. • Graphs may be different in terms of scale so resize as needed as seen in the graph. Practice Problem #1 • A class of fourth graders take a diagnostic reading test, and the scores are reported by reading the grade level. The 5 number summaries for the 14 boys and 11 girls are shown: • Boys: 2.0, 3.9, 4.3, 4.9, 6.0 • Girls: 2.8, 3.8, 4.5, 5.2, 5.9 • Find which group has the highest score • Which group had the greatest range • Which group had the greatest interquartile range • Which group’s scores appear to be more skewed? Explain. • If the mean reading level for boys was 4.2 and for girls was 4.6, what is the overall mean for the class? Practice Problem #1 cont • • • • • • • • Boys: 2.0, 3.9, 4.3, 4.9, 6.0 Girls: 2.8, 3.8, 4.5, 5.2, 5.9 A. The boys have the highest score 6.0 B. Boys have greatest range (6.0-2.0) opposed (5.9-2.8) C. Greatest IQR Girls (5.2-3.8) opposed (4.9-3.9) D. Boys more skewed, larger range skewed more to right with IQRs not quite equal E. Girls median and upper quartile better F.[14(4.2) +11(4.6)]/25 =4.38 Practice Problem #2 • The National Center for Education Statistics reported 1999 average mathematics achievement scores for eighth graders in 38 nations. Singapore led the group, with an average score of 604, while South Africa had the lowest average of 275. The United States scored 502. The average scores for each nation are given below: • 604 587 585 582 579 558 540 534 532 531 530 526 525 520 520 519 511 505 502 496 491 482 479 476 472 469 467 466 448 447 429 428 422 403 392 345 337 275 • A. Find the 5 number summary, IQR, mean, and standard deviation of averages • B. Write a brief summary of eighth graders worldwide be sure to comment on the US. Practice Problem #2 cont • 604 587 585 582 579 558 540 534 532 531 530 526 525 520 520 519 511 505 502 496 491 482 479 476 472 469 467 466 448 447 429 428 422 403 392 345 337 275 • A. 5 numbered summary: 275, 448, 499, 531, 604. IQR: 83 Mean: 487.21 SD: 71.36 • B. The distribution is unimodal skewed left, there is one outlier, an average score of 275. The median was 499, while the mean was 487.21 slightly below the median score. The middle 50% of the nations scored between 448 and 531 for an IQR of 83.