MATH 215 - Midterm Cheat sheet Athabasca University Question 1: Observations from the first 25 days of the year showed that the following number of class boundary midpoint frequency cum freq relative frequency pedestrians used a particular cross walk on each day: 15-34 14.5 -34.5 15+34/2=24.5 4 4 4/25=.16 19 132 25 78 62 101 84 32 99 71 43 56 64 81 59 67 43 97 105 91 31 66 76 87 125 35-54 34.5 -54.5 35+54/2=44.5 2 6 2/25=.08 a. construct a frequency distribution using lower limit of 15 and class width of 20 — indicate class 55-74 54.5 -74.5 55+74/2=64.5 7 13 7/25=.28 limits, boundaries, midpoints, frequencies and cumulative frequencies 75-94 74.5 -94.5 75+94/2=84.5 6 19 6/25=.24 b. create relative frequency polygon 95-114 94.5 -114.5 95+114/2=104.5 4 23 4/25=.16 c. create a percentage ogive — cumulative relative frequency 115-134 114.5 -134.5 115+134/2=124.5 2 25 2/25=.08 d. calculate approx. value of 30th percentile and what does it mean: k = percentile + n = sample size organize data from least to greatest — k x n / 100 = P30 — 30 x = 750/100 = 7.5 (round up = 8) find the 8th data — 30 percentile = 30% of the data is below that percentile 3. An average glass of milk contains 115 grams of calcium with a standard deviation of 8 grams. Using Chebyshev’s theorem, construct an interval that contains the calcium content of at least 50% of the glasses of milk. mean = 115 — standard deviation = 8 — percentage = 50 2. inferential or descriptive I D — in my son’s grade 3 class, 25% of the kids are under 8 and 75% are 8 or older I D — a comparison between placebo and drug treatments has determined that drug x produced a %5 reduction in the severity of symptoms of alzheimers patients I D — a non-leap year consists of 525600 minutes I D — by examining the browsing habits of individuals last week, a well -known internet retailer has estimated that 13% of those people who visit their website will make a purchase 4. the monthly cell phone bills for 100 senior citizens are summarized in the following frequency distribution table cell phone bill ($) frequency midpoint mf m2 m2f .50 = 1 - 1/K2 50 = 100 - 1 100 100 K2 1 = 100 - 50 K2 100 100 100 = 50K2 50 50 2 = K2 1.414213562 = k 1.41 = k a = 115 - 1.41 x 8 = 103.72 b = 115 + 1.41 x 8 = 126.28 At least 50% of the glasses contain a calcium level between 103.72 grams and 126.28 grams a. using the frequency distribution table calculate the mean cell phone bill amount mean = ∑mf = 4840 = 48.40 The mean cell phone bill is $48.40 N 100 b. using the frequency distribution table and treating the data as population data calculate the standard deviation σ2 = ∑m2f - (∑mf)2 247300 - (4840)2 5c. cv = s x 100% = 8.2686 x 100% = 39.3695% x 21 d. calculate the quartiles and IQR: 0 9 14 15 16 18 19 20 21 | 21 25 26 26 27 28 30 31 32 IQR = 26 -16=11 lowest value within inner fence is 0 11x1.5 = 16.5 highest value within inner fence is 32 16-16.5 = -.5 there are no outliers in this data 27+16.5 = 43.5 f. what is an advantage of using a box-and-whisker plot compared to a stem and leaf visualize center, spread and skewness. detect outliers — more information g. what is the percentile rank of a teen who smokes 31 per day — what does this say? number of data below 31 = 16 x 100 = 88.89 (round up) — 89% number of observations 18 89% of the teens smoke than than or equal to 31 per day h. true or false T F — if the sample of teens is selected in a way that ensures that each one has an equal chance of being selected, this is a random sample T F — if the sample is selected in a way that ensures that 9 of them are selected from the male and the other nine from the females, this is systematic random sampling 10 to less than 20 1 20+10/2 = 15 1x15 = 15 225 225x1 = 225 20 to less than 30 9 30+20/2 = 25 9x25 = 225 625 625x9 = 5625 30 to less than 40 10 30+40/2 = 35 10x35 = 350 1225 1225x10 = 12250 40 to less than 50 26 40+50/2 = 45 26x45 = 1170 2025 2025x26 = 52650 50 to less than 60 43 50+60/2 = 55 43x55 = 2365 3025 3025x43 = 130075 60 to less than 70 11 60+70/2 = 65 11x65 = 715 4225 4225x11 = 46475 Total 100 ∑mf = 4840 ∑m2f = 247300
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