Descriptive Statistics Histogram/3M/frequency Table [124 .
descriptive statistics histogram/3M/frequencytable [124 marks]1a. A group of students were asked how long they spend practising mathematics during the week. The results are shown in thefollowing table.[3 marks]It is known that 35 a 52 .Write downi)the modal class;ii)the mid-interval value of the modal class;iii)the class in which the median lies.Markschemei)3 t 4(A1)(C1)Note: Accept equivalent notation: [3, 4) or [3, 4[.ii)3.5(A1)(ft)(C1)Note: Follow through from part (a)(i).iii)2 t 3(A1)(ft)(C1)Note: Follow through from part (a)(i), for consistent misuse of inequality. Accept equivalent notation: [2, 3) or [2, 3[.1b. For this group of students, the estimated mean number of hours spent practising mathematics is 2.69.[3 marks]Calculate the value of a .Markscheme3.5 0.5 30 1.5 a 2.5 52 3.5 43 4.535 30 a 52 43 2.69(M1)(A1)(ft)Notes: Award (M1) for substitution into mean formula and equating to 2.69, (A1)(ft) for correct substitutions. Follow through from theirmid-interval value in part (a)(ii).(a ) 40(A1)(ft)(C3)Note: The final (A1)(ft) is awarded only if a is an integer and 35 a 52. Follow through from part (a)(ii).
Daniel grows apples and chooses at random a sample of 100 apples from his harvest.He measures the diameters of the apples to the nearest cm. The following table shows the distribution of the diameters.2a. Using your graphic display calculator, write down the value of(i)the mean of the diameters in this sample;(ii)the standard deviation of the diameters in this sample.[3 marks]Markscheme(i)6.76 (cm)(G2)Notes: Award (M1) for an attempt to use the formula for the mean with a least two rows from the table.(ii)1.14 (cm) (1.14122 (cm))(G1)2b. Daniel assumes that the diameters of all of the apples from his harvest are normally distributed with a mean of 7 cm and astandard deviation of 1.2 cm. He classifies the apples according to their diameters as shown in the following table.[3 marks]Calculate the percentage of small apples in Daniel’s harvest.MarkschemeP(diameter 6.5) 0.338 (0.338461)(M1)(A1)Notes: Award (M1) for attempting to use the normal distribution to find the probability or for correct region indicated on labelleddiagram. Award (A1) for correct probability.33.8(%)(A1)(ft)(G3)Notes: Award (A1)(ft) for converting their probability into a percentage.2c.Daniel assumes that the diameters of all of the apples from his harvest are normally distributed with a mean of 7 cm and astandard deviation of 1.2 cm. He classifies the apples according to their diameters as shown in the following table.Of the apples harvested, 5% are large apples.Find the value of a.[2 marks]
MarkschemeP(diameter a) 0.05(M1)Note: Award (M1) for attempting to use the normal distribution to find the probability or for correct region indicated on labelled diagram.a 8.97 (cm) (8.97382 )2d.(A1)(G2)Daniel assumes that the diameters of all of the apples from his harvest are normally distributed with a mean of 7 cm and astandard deviation of 1.2 cm. He classifies the apples according to their diameters as shown in the following table.[2 marks]Find the percentage of medium apples.Markscheme100 (5 33.8461 )(M1)Note: Award (M1) for subtracting “5 their part (b)” from 100 or (M1) for attempting to use the normal distribution to find the probabilityP (6.5 diameter their part (c)) or for correct region indicated on labelled diagram. 61.2(%) (61.1538 (%))(A1)(ft)(G2)Notes: Follow through from their answer to part (b). Percentage symbol is not required. Accept 61.1(%) (61.1209 (%)) if 8.97 used.2e.Daniel assumes that the diameters of all of the apples from his harvest are normally distributed with a mean of 7 cm and astandard deviation of 1.2 cm. He classifies the apples according to their diameters as shown in the following table.This year, Daniel estimates that he will grow100 000 apples.Estimate the number of large apples that Daniel will grow this year.Markscheme100 000 0.05(M1)Note: Award (M1) for multiplying by 0.05 (or5%). 5000(A1)(G2)[2 marks]
In a particular week, the number of eggs laid by each hen on a farm was counted. The results are summarized in the following table.3a.State whether these data are discrete or continuous.[1 mark]Markschemediscrete3b.(A1)(C1)[2 marks]Write down(i)the number of hens on the farm;(ii)the modal number of eggs laid.Markscheme3c.(i)60(A1)(ii)5(A1)(C2)[3 marks]Calculate(i)the mean number of eggs laid;(ii)the standard deviation.Markscheme1 4 2 7 3 12 60(i)(M1)Notes: Award (M1) for an attempt to substitute into the “mean of a set of data” formula, with at least three correct terms in thenumerator.Denominator must be 60.Follow through from part (b)(i), only if work is seen. 4.03 (4.03333 )(A1)Notes: Award at most (M1)(A0) for an answer of 4 but only if working seen.(ii)1.54 (1.53803 )(A1)(C3)A survey was carried out on a road to determine the number of passengers in each car (excluding the driver). Thetable shows the results of the survey.4a. State whether the data is discrete or continuous.[1 mark]
Markschemediscrete(A1)(C1)[1 mark][1 mark]4b. Write down the mode.Markscheme0(A1)(C1)[1 mark]4c. Use your graphic display calculator to find(i)the mean number of passengers per car;(ii)the median number of passengers per car;(iii)the standard deviation.[4 marks]Markscheme(i)1.47(1.46666.)(A2)Note: Award (M1) for176120seen.Accept1 or2 as a final answer if1.4666 or1.47 seen.(ii)1.5(A1)(iii)1.25(1.25122.)(A1)(C4)[4 marks]The number of passengers in the first ten carriages of a train is listed below.6,8,6,3,8,4,8,5p, ,pThe mean number of passengers per carriage is 5.6.5a. Calculate the value of p.[2 marks]
Markscheme48 2p10 5.6(M1)Notes: Accept equivalent forms. Award (M1) for correct substitutions in mean formula.4(A1)(C2)5b. Find the median number of passengers per carriage.[2 marks]MarkschemeCorrectly rearranging the list with their p5.5(A1)(ft)(M1)(C2)Note: Follow through from their value ofp in part (a).5c. If the passengers in the eleventh carriage are also included, the mean number of passengers per carriage[2 marks]increases to 6.0.Determine the number of passengers in the eleventh carriage of the train.Markscheme56 x11 6.0(M1)Notes: Accept equivalent forms. Award (M1) for correct substitutions in mean formula.OR48 2 their part (a) x1110(A1)(ft)(M1)(C2)Note: Follow through from their answer to part (a).The table below shows the frequency distribution of the number of dental fillings for a group of25 children.6a. Find the value ofq.[2 marks]
Markschemeq 25 (4 3 8 4 1)(M1)Note: Award (M1) for subtraction from25 of all values from the table. 5(A1)(C2)[2 marks]6b. Use your graphic display calculator to find[4 marks](i) the mean number of fillings;(ii) the median number of fillings;(iii) the standard deviation of the number of fillings.Markscheme(i)2.2(A2)(ft)(C2)Note: Award (M1) for use of mean formula with correct substitution. Follow through from part (a), irrespective ofwhether working is shown.(ii)2(A1)(C1)(iii)1.39(A1)(ft)(C1)Note: Follow through from part (a), irrespective of whether working is shown. Award(A1) for1.38.[4 marks]80 matches were played in a football tournament. The following table shows the number of goals scored in allmatches.7a. Find the mean number of goals scored per match.[2 marks]Markscheme0 16 1 22 2 19 80(M1)Note: Award (M1) for substituting correct values into mean formula.1.75(A1)(C2)[2 marks]7b. Find the median number of goals scored per match.[2 marks]
MarkschemeAn attempt to enumerate the number of goals scored.2(A1)(M1)(C2)[2 marks]7c. A local newspaper claims r attempt to order raw data (if frequency table not used) or(M1) halfway between 10th and 11th result.[2 marks]
12d. One student is chosen at random from the group.[1 mark]Find the probability that this student obtained either grade4 or grade5.Markscheme720(0.35, 35%)(A1)(ft)(C1)[1 mark]The distribution of the weights, correct to the nearest kilogram, of the members of a football club is shown in thefollowing table.13a. On the grid below draw a histogram to show the above weight distribution.[2 marks]Markscheme(A1)(A1)(C2)Notes: (A1) for all correct heights, (A1) for all correct end points (39.5,49.5 etc.).Histogram must be drawn with a ruler (straight edge) and endpoints must be clear.Award (A1) only if both correct histogram and correct frequency polygon drawn.[2 marks]13b. Write down the mid-interval value for the40 49 interval.[1 mark]
Markscheme44.5(A1)(C1)Note: If (b) is given as45 then award(b)45 (A0)(c)58.8 kg (M1)(A1)(ft) or (C2)(ft) if no working seen.(d)8.44 (C1)[1 mark]13c. Find an estimate of the mean weight of the members of the club.[2 marks]MarkschemeUnit penalty (UP) applies in this question.Mean 44.5 6 54.5 18 42(M1)Note: (M1) for a sum of frequencies multiplied by midpoint values divided by42. 58.3 kg(A1)(ft)(C2)Note: Award (A1)(A0)(AP) for58.Note: If (b) is given as45 then award(b)45 (A0)(c)58.8 kg (M1)(A1)(ft) or (C2)(ft) if no working seen.(d)8.44 (C1)[2 marks]13d. Write down an estimate of the standard deviation of their weights.MarkschemeStandard deviation 8.44(A1)(C1)Note: If (b) is given as45 then award(b)45 (A0)(c)58.8 kg (M1)(A1)(ft) or (C2)(ft) if no working seen.(d)8.44 (C1)[1 mark][1 mark]
14. Complete the following table of values for the height and weight of seven students.[4 marks]MarkschemeMode 171(A1)Median 148, 151, 158, 163, 171, 171, 184 163(A1)Mean 64.7(A1)Standard deviation 13.3(A1)(C4)Note: If both mean and standard deviation given to 2 significant figuresMean 65, (A0)(AP)Standard deviation 13 (A1)(ft) ((AP) already deducted).[4 marks]The table below shows the number of words in the extended essays of an IB class.15a. Draw a histogram on the grid below for the data in this table.[3 marks]
Markscheme(A3)(C3)Notes: (A3) for correct histogram, (A2) for one error, (A1) for two errors, (A0) for more than two errors.Award maximum (A2) if lines do not appear to be drawn with a ruler.Award maximum (A2) if a frequency polygon is drawn.[3 marks]15b. Write down the modal group.[1 mark]MarkschemeModal group 3800 w 4000(A1)(C1)[1 mark]15c. The maximum word count is[2 marks]4000 words.Write down the probability that a student chosen at random is on or over the word count.MarkschemeProbability 335(0.0857, 8.57%)(A1)(A1)(C2)Note: (A1) for correct numerator (A1) for correct denominator.[2 marks]
The following histogram shows the weights of a number of frozen chickens in a supermarket. The weights are groupedsuch that1 weight 2,2 weight 3 and so on.16a. Find the total number of chickens.[1 mark]Markscheme96(A1)(C1)[1 mark]16b. Write down the modal group.[1 mark]Markscheme3 weight 4 kg . Accept3 4 kg (A1) (C1)[1 mark]16c. Gabriel chooses a chicken at random.[2 marks]Find the probability that this chicken weighs less than4 kg.MarkschemeFor adding three heights or subtracting14 from96 (M1)8296(0.854 or[2 marks]41,4885.4%) (ft) from (b).(A1)(ft)(C2)
A survey was conducted of the number of bedrooms in208 randomly chosen houses. The results are shown in the following table.17a. State whether the data is discrete or continuous.[1 mark]MarkschemeDiscrete(A1)(C1)[1 mark]17b. Write down the mean number of bedrooms per house.[2 marks]MarkschemeFor attempting to find fx/ f (M1)2.73(A1)(C2)Note: for (b) and (c), if both mean and standard deviation given to 2 significant figures.Award (C1)(C0)(AP) for2.7. Award (A1)(ft) for1.3 ((AP) already deducted).[2 marks]17c. Write down the standard deviation of the number of bedrooms per house.[1 mark]Markscheme1.34(A1)(C1)Note: for (b) and (c), if both mean and standard deviation given to 2 significant figures.Award (C1)(C0)(AP) for2.7. Award (A1)(ft) for1.3 ((AP) already deducted).[1 mark]17d. Find how many houses have a number of bedrooms greater than one standard deviation above the mean.MarkschemeAttempt to find their mean their standard deviation (can be implied)(M1)23, (ft) their mean and standard deviation.(A1)(ft)[2 marks](C2)[2 marks]
The figure below shows the lengths in centimetres of fish found in the net of a small trawler.18a. Find the total number of fish in the net.[2 marks]MarkschemeTotal 2 3 5 7 11 5 6 9 2 1 (M1)(M1) is for a sum of frequencies. 51(A1)(G2)[2 marks]18b. Find (i) the modal length interval,[5 marks](ii) the interval containing the median length,(iii) an estimate of the mean length.MarkschemeUnit penalty (UP) is applicable where indicated in the left hand column.(i) modal interval is 60 – 70Award (A0) for 65(A1)(ii) median is length of fish no. 26,also 60 – 70(M1)(A1)(G2)Can award (A1)(ft) or (G2)(ft) for 65 if (A0) was awarded for 65 in part (i).(iii) mean is2 25 3 35 5 45 7 55 .51(UP) 69.5 cm (3sf)(M1)(A1)(ft)(G1)Note: (M1) is for a sum of (frequencies multiplied by midpoint values) divided by candidate’s answer from part (a).Accept mid-points 25.5, 35.5 etc or 24.5, 34.5 etc, leading to answers 70.0 or 69.0 (3sf) respectively. Answers of 69.0,69.5 or 70.0 (3sf) with no working can be awarded (G1).[5 marks]18c. (i) Write down an estimate for the standard deviation of the lengths.(ii) How many fish (if any) have length greater than three standard deviations above the mean?[3 marks]
MarkschemeUnit penalty (UP) is applicable where indicated in the left hand column.(UP) (i) standard deviation is 21.8 cm(G1)For any other answer without working, award (G0). If working is present then (G0)(AP) is possible.(ii)69.5 3 21.8 134.9 120no fish(M1)(A1)(ft)(G1)For ‘no fish’ without working, award (G1) regardless of answer to (c)(i). Follow through from (c)(i) only if method isshown.[3 marks]18d. The fishing company must pay a fine if more than 10% of the catch have lengths less than 40cm.[2 marks]Do a calculation to decide whether the company is fined.Markscheme5 fish are less than 40 cm in length,(M1)Award (M1) for any of5,5146,510.098 or 9.8%, 0.902, 90.2% or 5.1 seen.hence no fine.(A1)(ft)Note: There is no G mark here and (M0)(A1) is never allowed. The follow-through is from answer in part (a).[2 marks]18e. A sample of 15 of the fish was weighed. The weight,W was plotted against length, L as shown below.Exactly two of the following statements about the plot could be correct. Identify the two correct statements.Note: You do not need to enter data in a GDC or to calculate r exactly.(i) The value of r, the correlation coefficient, is approximately 0.871.(ii) There is an exact linear relation between W and L.(iii) The line of regression of W on L has equation W 0.012L 0.008 .(iv) There is negative correlation between the length and weight.(v) The value of r, the correlation coefficient, is approximately 0.998.(vi) The line of regression of W on L has equation W 63.5L 16.5.[2 marks]
Markscheme(i) and (iii) are correct.(A1)(A1)[2 marks]Consider the frequency histogram for the distribution of the time,t , in minutes of telephone calls that Helen made last week.19a. Complete the frequency table for this distribution.[2 marks]Markscheme(A2)(C2)Note: Award (A2) for four correct entries, (A1) for three correct entries, (A0) otherwise.19b. Write down the modal class.Markscheme0 t 5(A1)(C1)[1 mark]
19c. Write down the mid interval value of the[1 mark]10 t 15 class.Markscheme12.5(A1)(C1)19d. Use your graphic display calculator to find an estimate for the mean time.[2 marks]Markscheme27530(M1)Note: Award (M1) for division with275 seen. 9.17 (9.16666 )(A1)(ft)(C2)Note: Follow through from their parts (a) and (c), irrespective of whether working is shown. International Baccalaureate Organization 2017International Baccalaureate - Baccalauréat International - Bachillerato Internacional Printed for Victoria Shanghai Academy
The table below shows the frequency distribution of the number of dental fillings for a group of children. Find the value of. 25 q. Markscheme (M1) Note: Award (M1) for subtraction from of all values from
To create our Histogram Bins, we start by entering the Column Header "Histogram Bins" into a blank column in our spreadsheet. In the first cell of this column, we enter our first bin value (our first interval). In this example our first bin value is 1 to 3. We then continue to enter the rest of our intervals into the Histogram Bins column.
Use the maximum and minimum data entries and the number of classes to find the class width, the lower class limits, and the upper class limits. 5) min 1, max 30, 6 classes . Use the given frequency distribution to construct a frequency histogram, a relative frequency histogram and a frequency polygon. 10) Height (in inches) Class Frequency, f
4. Descriptive statistics Any time that you get a new data set to look at one of the first tasks that you have to do is find ways of summarising the data in a compact, easily understood fashion. This is what descriptive statistics (as opposed to inferential statistics) is all about. In fact, to many people the term "statistics" is
Descriptive plots: histogram, boxplot, Scatter Plot 2. Descriptive statistics: location, spread 7. Barplot for Categorical Data Useful to describe the distribution of values for a categorical variable 8 Proportion of cases versus stage of can
descriptive statistics available, many of which are described in the preceding section. The example in the above dialog box would produce the following output: Going back to the Frequencies dialog box, you may click on the Statistics button to request additional descriptive statistics. Click
Descriptive Analysis Univariate Bivariate Multi-variate Descriptive Analysis Frequency Table Simple Means (Mode, Median) Cross-tabulations Histogram Statistics of the Mean Statistics of Deviation Useful Analysis Appropriate Scale Level Nominal Ordinal Metric 3. Data Analysis 15
Nonparametric density estimate: a histogram I break data into bins and use relative frequency within each bin I Problem: a histogram is a step function, even if data are continuous Smooth nonparametric density estimate: kernel density estimate. I smooths a histogram in two way
Statistics for Engineers 4-1 4. Introduction to Statistics Descriptive Statistics Types of data A variate or random variable is a quantity or attribute whose value may vary from one unit of investigation to another. For example, the units might be headache sufferers and the variate might be the time between taking an aspirin and the headache ceasing. An observation or response is the value .
