Chapter 9. There are 7 out of 50 measurements that are greater than or equal to 5.1; therefore, the fraction of the

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Kelley Davidson
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1 Pratie questions 6 1 a y i = 6 µ = = 1 i = 1 y i µ i = 1 ( ) = 95 = s n x i f i n n µ = = = f 11+ n 11+ n i 7 + n = 5 + n = 6n n = a Time (minutes) Frequeny Frequeny Time There are 7 out of 5 measurements that are greater than or equal to 5.1; therefore, the fration of the measurements less than 5.1 is: = 86. = 86%. 5 There are 5 piees of ata, so, to etermine the meian, we nee to fin the 5th an 6th oservations. We notie that these two oservations are within the interval.6.1; therefore, the meian is approximately.9. We an input the time as a sequene y using the sequene list feature on a GDC. The mean value is.68, whilst the stanar eviation is 1.11, orret to three signifiant figures. 1

e 5 Cumulative frequeny 1 5 6 Time f Estimates for the minimum an maximum values are 1.6 an 6.6 respetively. The first an thir quartiles orrespon to the umulative frequenies of 1.5 an 7.

2 e 5 Cumulative frequeny Time f Estimates for the minimum an maximum values are 1.6 an 6.6 respetively. The first an thir quartiles orrespon to the umulative frequenies of 1.5 an 7.5 respetively; therefore, an estimate for the first quartile is.15 an the thir quartile is.65. An estimate for the meian (whih orrespons to a umulative frequeny of 5) is.9. a The meian an the IQR woul est represent the ata, sine the ata is skewe to the right an there are a few outliers on the right. Firstly, we are going to rea the frequenies from the histogram an input them into the frequeny istriution tale on our GDC. Finally, we apply the statistial alulation for one variale. The mean value is 68.6 an the stanar eviation is 56.. Sine we have groupe ata, the enpoints of our intervals will e 15, 5, 5,..., an so on. On a alulator, we an use the aing a numer to the list feature. Alternatively, we an alulate the umulative frequenies from the List menu.

3 There are 6 ities, so, to estimate the meian, we nee to raw a horizontal line from. To fin the lower an upper quartiles, we nee to raw horizontal lines from 115 an 5 respetively. For a etter estimation, we will use the zoom ox feature an turn the gris on. So, the meian is aout 5. The first quartile is aout an the thir quartile is aout 8. Therefore, the IQR is aout 8 = 5. e f There are a few outliers to the right. The outliers are those points whih are over Q + 1.5IQR, i.e = 158, whih gives us 5 ities from the histogram. The ata is skewe to the right with quite a few outliers to the right (16 an aove.) The ata is also imoal, with the moal values eing an. 5 a It appears that Spain proues oth the most expensive (estimate 15 per ase) an the heapest (estimate 55 per ase) re wine. Re wines are generally more expensive in Frane as we an see that the meian prie is the highest; the minimum value in Frane is also the highest, ut the upper 5% of wines are also within a very small range of approximately 1 per ase. It appears that the wines are, on average, more expensive in Frane, where the pries are skewe towars the higher en. In Spain, you an fin a higher perentage of heaper wine than in the other two ountries, ut you also fin the most expensive wines on the market; so Spain has the wiest range of pries. Italy seems to have the most symmetrial istriution of wine pries. 6 a The mean value of the ata is 5.6 an the stanar eviation is 7.6, oth given orretly to three signifiant figures. The meian value is 51.. The upper an lower quartiles are 9.9 an 5.6 respetively; therefore, the IQR is.65, all orret to three signifiant figures. Sine there is one outlier (11.7), the mean value is more influene y it than the meian value an IQR. 7 a The istriution is not symmetri sine the meian is not the mipoint etween the minimum an maximum value, nor is it the mipoint etween the first an the thir quartile. The outliers lie 1.5IQR further to the left of the lower quartile an 1.5IQR further to the right of the upper quartile. If there are any outliers, they will lie elow 7 an aove 99, whih is outsie of the given range; therefore, there are no outliers.

4 y 1 x Aitionally to the esription provie in part a, we an say that the ata is skewe to the left. 8 a From the graph, we an estimate that the meian holesterol level (5% of the umulative perentage) is 5. The first an thir quartiles are estimate from 5% an 75% of the umulative perentage; our estimations are 1 an 6 respetively. The 9th an 1th perentiles are estimate from 9% an 1% of the umulative perentage; our estimations are an 19 respetively. Using our answers from part, the estimate IQR will e 5. Sine patients have een stuie, the numer of patients in the mile 5%, ranging from 1 to 6, is 1. y 1 x e We notie that the ata is skewe to the right a it, with more outliers on the right sie, sine the outliers lie outsie of the interval 15 to 5. From the umulative frequeny graph, we rea that there are almost 1 patients who have a holesterol level greater than 5 mg/l an only a few patients with a level less than 15 mg/l. 9 a Spee Frequeny Frequeny ensity 8 6 Spee

5 x i f i x i f i x i f i Σ x i f i i µ = 8.1 s =.99 s = e In orer to estimate the meian, we nee to raw the umulative frequeny iagram. We will o this using a alulator. An estimate of the meian is 7. Q 1 is an Q is ; therefore, the IQR is 8. f Sine Q is an 1.5IQR is 1, there is a possile outlier, whih is the largest oservation of 5. 1 a Classes Frequeny Using Autograph software with the groupe ata, we otain the following results: The mean value is an the stanar eviation is.79. The meian value is The first an thir quartiles are an 19.5 respetively; therefore, the IQR is By looking at the frequeny istriution tale, we an surmise that there may e some outliers to the left, so we nee to alulate Q 1 1.5IQR = 156. From the tale, we an see that we have one outlier. 5

6 5 f 1 Gasoline (l) µ ± s n = ±. 79 [ 1618, 8] e Germany will efinitely e an outlier sine Q + 1.5IQR = 77 < 758; therefore, it will influene the mean an stanar eviation. Conversely, the meian, the first an thir quartiles, an the IQR will not hange muh a x i = 6 µ = 9. 6 minutes i = µ New = = minutes 1 a Marks < 1 < < < < 5 < 6 < 7 < 8 < 9 < 1 Numer of aniates Cumulative frequeny Marks i By looking at the graph, we estimate that the meian sore (whih orrespons to the umulative frequeny of 1) is 6. We raw a vertial line from 5 on the Marks axis an reah the umulative frequeny urve at the point at whih the umulative frequeny is aout 5. Therefore, 5 aniates ha to retake the exam. i The highest soring 15% orrespons to the highest results; therefore, we raw a horizontal line from 17 on the Cumulative frequeny axis an reah the urve at the point at whih the numer of marks is aout 6. Hene, a istintion will e aware if 6 or more marks are sore on the test µ = = 17. m, orret to the nearest entimetre a x i = µ = = 1 5 i = 1 5 i = 1 ( ) = = = x i m 65 s n

15 a To alulate an estimate for the mean, we will take the mipoints of the intervals (15, 5, 75, an so on) an the orresponing frequenies. So, an estimate for the mean of the waiting times is 97.

7 15 a To alulate an estimate for the mean, we will take the mipoints of the intervals (15, 5, 75, an so on) an the orresponing frequenies. So, an estimate for the mean of the waiting times is 97. seons. (Note: Even though we oul have use only the mean feature from the List menu, we use the whole statistis alulation sine there is a hane that we will nee further statistis in the following parts of the prolem.) Waiting time (seons) < < 6 < 9 < 1 < 15 < 18 < 1 < Cumulative frequeny Cumulative frequeny Waiting time (s) To fin the three estimations aske for, we nee to raw a horizontal line at 5 for the meian an at 5 an 75 for the quartiles. An estimation of the mean value is 87, whilst the lower an upper quartiles are 65 an 11 respetively. 16 a Taking our reaings from the histogram: i There are 1 plants that have a length etween 5 an 6 m. To fin the numer of trees that have a length etween 7 an 9 m, we nee to a two frequenies: =. As in question 15, we again take the mipoints of the intervals (15, 5, 5,..., 95) an the orresponing frequenies. x i f i We input the two lists into a GDC an get the following results. So, an estimate for the mean is 6, an the stanar eviation is.5. 7

The ata is skewe to the left; therefore, the meian an the mean are ifferent values. Sine there are 8 plants, the meian orrespons to the umulative frequeny of.

17 a Again, we will use the mipoints of the intervals (8.5, 87.5, 9.5,..., an so on) an the orresponing frequenies. We put the two lists into our GDC an otain an estimation of the stanar eviation.

8 The ata is skewe to the left; therefore, the meian an the mean are ifferent values. Sine there are 8 plants, the meian orrespons to the umulative frequeny of. Sine the values of the umulative frequenies are symmetrial aroun, for 6 an 8 for 7, we an estimate the umulative frequeny of woul e 65, i.e. the meian is less than or equal to 65 m. 17 a Again, we will use the mipoints of the intervals (8.5, 87.5, 9.5,..., an so on) an the orresponing frequenies. We put the two lists into our GDC an otain an estimation of the stanar eviation. So, an estimate for the stanar eviation of the weights is 7.5. Weight (W) W < 85 W < 9 W < 95 W < 1 W < 15 W < 11 W < 115 Numer of pakets Cumulative frequeny Weight (g) i An estimate for the meian (whih orrespons to the umulative frequeny of ) is 97. An estimate for the upper quartile (whih orrespons to the umulative frequeny of 6) is 11. ( W1 W)+ ( W W)+ ( W W) ( W8 W) = W1 + W + W W8 8 ( ) W W1 + W + W W8 = ( W1 + W + W W8 ) 8 8 e There are 71 pakets that satisfy the onition 85 < W 11. There are pakets that satisfy the onition 1 < W 11. Therefore, the proaility is: P( E) =. 8, orret to three signifiant figures a We will use the mipoints of the intervals (65, 75, 85,..., an so on) an the orresponing frequenies. We put the two lists into our GDC an alulate the mean. = So, the mean spee is 98. km h 1, orret to three signifiant figures. 8

9 i To fin the value of a we an either a 7 (the frequeny of the spee interval 9 1) to the previous umulative frequeny, 95; or sutrat 71 (the frequeny of the spee interval 1 11) from the next umulative frequeny, 6. In oth ases we get the same value: a = 165. In a similar way, we fin the value of : = = 75. Cumulative frequeny Spee (km/h) i We raw a vertial line from v = 15 until we reah the umulative frequeny urve. This gives us an estimate for the umulative frequeny of. So, there are 1 ars that will exee the spee of 15 km h 1 an 1 P = =.....%. If 15% of the ars exee this spee, then 85% o not exee that spee. 85% of is 55, so we raw a horizontal line from y = 55 until we reah the umulative frequeny urve. This gives us an estimate of a spee of 115 km h a i We take a horizontal line aross from 1 on the vertial axis until it touhes the graph. From that point, we take a vertial line own to the horizontal axis an rea off the value. An estimate for the meian fare is $. We take a vertial line up from 5 on the horizontal axis until we reah the graph. From that point, we take a horizontal line aross to the vertial axis an rea off the value. An estimate for the numer of as in whih the fare taken is $5 or less is 158. % of the as is. = 8. So, we take a horizontal line from 8 on the vertial axis until we reah the graph an then we estimate the x-oorinate, whih is. Therefore, the fare is $. To fin the numer of kilometres, we nee to ivie the fare y.55 (whih is the fare per kilometre for istane travelle). Therefore, the istane travelle is km. If the istane travelle is 9 km, the river will earn = 95. ollars. We will use the graph to estimate the numer of as that will earn less than 9.5 ollars: there are 18, an therefore there are 16 as that will earn more than that. So, the perentage of the as that travel more than 9 km is: 16 = 8. = 8%. Sine the three numers are given in orer of magnitue, a < <, we know that the mile one (the meian) is 11, so = 11. Given that the range is 1, we know that the ifferene etween the minimum an maximum value is: a = 1. Sine the mean value is 9, we an estalish another equation in terms of a an : a = 9 a + = 16. Solving these two equations (using the elimination metho), we get: a = 6 a =. So, finally, = 1. 9

10 1 a Cumulative frequeny Prie ( thousans of $) By using horizontal lines at 5 an 75, we estimate the values of Q 1 an Q as $1 an $5 respetively. Hene, the IQR is $1. To fin the frequenies a an we nee to sutrat two suessive umulative frequenies: fi = i i 1. Therefore, a = 9 87 = 7, an = 1 9 = 6. We input the mipoints of the intervals in the first list an the orresponing frequenies in the seon list: So, an estimate of the mean selling prie is $199. e i An estimate of the umulative frequeny for $5 is 9; therefore, there are aout eight houses that an e esrie as De Luxe. Out of eight De Luxe houses, six were sol for $ ; therefore, the proaility that oth selete houses have a selling prie more than $ is: P( E) = = a i To mark the meian, we raw the horizontal line y = until it hits the graph, at whih point we raw a vertial line own to the Diameter axis. An estimate for the meian is mm. To mark the upper quartile, we raw the horizontal line y = 6 until it hits the graph, at whih point we raw a vertial line own to the Diameter axis. An estimate for the upper quartile is mm. The interquartile range is: IQR = 1 = 1 mm. a In this question we an aept an error of ± stuents. We nee to rea the umulative frequenies at the enpoints of the intervals an then sutrat the suessive ones to otain the frequenies. For the umulative frequeny is 7, an therefore the orresponing frequeny is 7 = 5. For 6 the umulative frequeny is 1, an therefore the orresponing frequeny is 1 7 = 68. For 8 the umulative frequeny is 18, an therefore the orresponing frequeny is 18 1 = 8. Mark (x) < x < < x < < x < 6 6 < x < 8 8 < x < 1 Numer of stuents

11 % of stuents is 8. The umulative frequeny of 8 orrespons to aout marks. So, the pass mark is aout %. a To fin the meian height we raw the horizontal line y = 6 until it hits the graph. We then estimate the x-oorinate of the point of intersetion. We estimate 18 m. For the lower an upper quartiles, we raw two horizontal lines: y = an y = 9. Then we estimate the x-oorinates of the points of intersetion. Therefore: Q = 175, Q = 189 IQR = = Sine the moal value is 11, we know that = = 11. Given that the range is 8, we an fin the value of a: 11 a = 8 a =. Finally, given that the mean value is 8, we an fin the remaining numer : = 8 5+ = = 7. 6 a We raw the vertial line x = until it hits the graph. We then estimate the y-oorinate of the point of intersetion as 1. So, the numer of stuents who sore marks or less is 1. There are 8 stuents, so the mile 5% is etween an 6 stuents. For a umulative frequeny of, the estimate mark is 55; whilst for 6, the estimate mark is 75. Hene, we say that the mile 5% of test results lie etween 55 an 75 marks: a = 55, =

Review Topic 4: Cubic polynomials

Review Topic 4: Cubic polynomials Review Topi : ui polynomials Short answer Fatorise Px ( ) = x + 5x + x- 9 into linear fators. The polynomial Px ( ) = x - ax + x- leaves a remainer of when it is ivie y ( x - ) an a remainer of - when