Priya - Statistics HW 2.3

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East West University *

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STA100A

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Statistics

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Apr 3, 2024

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2.60 The number of families is eight. So, n = 8 Sum of x is. ∑x = 1+2+1+3+2+1+5+3 = 18 So mean of X will be = 18 / 8 = 2.25 Hence, the mean of x will be = 2.25 --------------------------------------------------------------------------------------------------------------------- 2.63 (a) The number of states is 6. So, n = 6 Let x shows the miles in each state so sum of x is, ∑x = 16+132+124+191+183+299 = 945 = 945 / 6 = 157.5 Hence, the mean number of miles in each state along I-64 is = 157.5 miles (b) I-64 intersects with nine other interstate highways, besides I-264 and I-270 at its end points so there are total 11 interchanges. Therefore number of sections is 11 so, n = 11 ∑x = 16+132+124+191+183+299 = 945 So mean of x will be = 945 / 10 = 94.5 Hence, the mean distance between interchanges with other interstate highway along I-64 is 94.5 miles ---------------------------------------------------------------------------------------------------------------------

2.67 Let us rearrange the heights in ascending order: 70, 72, 73, 74, 76 Median is the middle data value of the ordered data. So, median height of a basketball team is 3rd data value of the data set. Hence, the required median is 73 inches. --------------------------------------------------------------------------------------------------------------------- 2.71 Mode of a data set is the data value which has largest frequency. So mode of the number of cars per apartment is 2. --------------------------------------------------------------------------------------------------------------------- 2.73 (a) The number of data values is six. So, n = 6 ∑x = 9+6+7+9+10+8 = 49 So, mean of x will be = 49/6 = 8.2 Hence, the mean of x will be = 8.2 arrange the data in ascending order: 6, 7, 8, 9, 9, 10 Median is the middle data value of the ordered data. Here number of values is even. So, median is average of 3rd and 4th value. Therefore, 8+9 / 2 Hence the median is 8.5 Mode of a data set is the data value which has largest frequency. So mode of the given numbers is 9. Highest value is 10 and lowest value is 6 so the midrange will be, Mid-range = Low value + high value / 2 = 6 + 10 / 2 = 8

Hence, the midrange is 8. (b) Following is the relationship between the mean, median, mode and midrange. --------------------------------------------------------------------------------------------------------------------- 2.74 (a) Here number of data values is five. So, n = 5 Sum of X is ∑x = 2+4+7+8+9 = 30 So, mean of x will be = 30 / 5 = 6 Hence the mean of X will be 6 (b) Let us arrange the data in ascending order: 2, 4, 7, 8, 9 Median is the middle data value of the ordered data. Here number of values is odd. So, median is 3rd value. Hence, median is 7. (c) Mode of a data set is the data value which has largest frequency. But in the data set all the values comes only once. So there is no mode of the data set. (d) Highest value is 9 and lowest value is 2 so the midrange will be, midrange = low value + high value / 2 = 2 + 9 / 2 = 5.5 Hence, the midrange is 5.5. --------------------------------------------------------------------------------------------------------------------- 2.77 (a) Let us find out mean age for the 2007 NASCAR drivers. Here the number of drivers is 10. So, n = 10 ∑x = 36+26+48+28+45+21+21+38+27+32 = 322 So mean of x will be,

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= 322 / 10 = 32.2 So, mean age for the 2007 NASCAR drivers is 32.2 (b) arrange the data in ascending order: 21, 21, 26, 27, 28, 32, 36, 38, 45, 48 Median is the middle data value of the ordered data. Here number of values is even. So median will be the average of 5th and 6th value. So, = 28 + 32 / 2 = 60 / 2 = 30 So, median age for the ten 2007 NASCAR drivers is 30. (c) Highest age in the given ages is 48 and lowest age is 21 so the midrange will be, midrange = low value + high value / 2 = 21 + 48 / 2 = 34.5 Hence, the midrange of age for the ten 2007 NASCAR drivers is 34.5 (d) Mode of a data set is the data value which has largest frequency. So mode age for the ten 2007 NASCAR drivers is 21 --------------------------------------------------------------------------------------------------------------------- 2.79 (a) The dot plot of the data is shown as: 0.02 0.02 0.02 0.02 0.03 0.03 0.03 0.04 0.04 0.04 0.05 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 Contact Lenses of critical feature X

The grouped frequency distribution of the data: The largest score is 0.042. The smallest score is 0.017. Range is 0.024. Class width is, Class width = Range / number of classes = 0.024 / 9 = 0.003 Pick a starting point: Here the starting point is lowest score, that starting point starts at 0.017 and using class width 0.003, the lower and upper limit of each class and their corresponding frequencies are shown in the table. Classes Frequency 0.017–0.020 2 0.020–0.023 2 0.023–0.026 6 0.026–0.029 4 0.029–0.032 0 0.032–0.035 6 0.035–0.038 2 0.038–0.041 1 0.041–0.042 2 The histogram for the frequency distribution is shown as:

(b) Let us find out mean for critical feature X. Here number of data values is 25 so n = 25 x represents the data value so sum of x is, ∑x = 0.714 So mean will be = 0.714 / 25 = 0.0286 So, mean for critical feature X is 0.0286. (c) Let us arrange the data in ascending order: 0.017, 0.019, 0.021, 0.022, 0.023, 0.023, 0.023, 0.023, 0.024, 0.024, 0.026, 0.027, 0.027, 0.027, 0.032, 0.032, 0.033, 0.033, 0.034, 0.034, 0.035, 0.035, 0.038, 0.041, 0.041 Median is the middle data value of the ordered data. Here number of values is odd. So median will be the 13th data value. So, median for the critical feature X is 0.027 (d) Highest data value is 0.041 and lowest data value is 0.017 so the midrange will be midrange = 0.014 + 0.017 / 2 = 0.029 Hence, the midrange for the critical feature X is 0.029 (e) Mode of a data set is the data value which has largest frequency. So mode for the critical feature X is 0.023. (f) Distribution is bimodal and has two parts. In this median (part c) fall relative to the distribution. Mean and mode do not give the correct measure for the center of distribution. (g) Distribution is bimodal and has two parts. One mode is near 0.023 and the second is near 0.032. --------------------------------------------------------------------------------------------------------------------- 2.85 (a) Following is the ranked data of the runs scored by the teams while playing at home: 3.57, 3.88, 3.95, 4.02, 4.16, 4.17, 4.33, 4.43, 4.46, 4.56, 4.59, 4.59, 4.62, 4.65, 4.73, 4.74, 4.79, 4.84, 4.93, 4.98, 5.07, 5.09, 5.09, 5.26, 5.32, 5.40, 5.60, 5.69, 5.76, 5.99 Mean: Here number of data values is 30 so

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n = 30 Let x represents the data value so sum of x is ∑x = 143.6 So mean of X will be = 143.6 / 30 = 4.7753 So, mean of the runs scored by the teams while playing at home is 4.78. Median: Median is the middle data value of the ordered data. Here number of values is even. So median will be the average of 15th and 16th data values. So, median will be = 4.73 + 4.74 / 2 = 4.735 So, median of the runs scored by the teams while playing at home is 4.735. Maximum value is 5.99 and minimum value is 3.57. Midrange: Highest data value is 5.99 and lowest data value is 3.57 so the midrange will be midrange = 5.99 + 3.57 / 2 = 4.78 Hence, the midrange of the runs scored by the teams while playing at home is 4.78. (b) Following is the ranked data of the runs scored by the teams while playing away: 4.00, 4.01, 4.02, 4.04, 4.05, 4.12, 4.15, 4.19, 4.26, 4.30, 4.31, 4.36, 4.38, 4.56, 4.62, 4.63, 4.65 4.67, 4.68, 4.72, 4.74, 4.74, 4.74, 4.78, 4.85, 5.00, 5.01, 5.01, 5.07, 5.14 Mean: Here number of data values is 30 so n = 30 Let x represents the data value so sum of x is ∑x = 135.80 So mean of x will be = 135.80 / 30 = 4.5266 So, mean of the runs scored by the teams while playing away is 4.53. Median: Median is the middle data value of the ordered data. Here number of values is even. So median will be the average of 15th and 16th data values. So, median will be

= 4.62 + 4.63 / 2 = 4.57 Hence, the midrange of the runs scored by the teams while playing away is 4.57. (c) Following table shows the statistics for both scores: Average Runs, Home Average runs, away Mean 4.78 4.53 Median 4.735 4.625 Maximum 5.99 5.14 Minimum 3.57 4.00 Midrange 4.78 4.57 Each measure except minimum value is greater for the runs scored in home. So we can conclude that the MLB is playing better at home in comparison when team played away. --------------------------------------------------------------------------------------------------------------------- 2.90 (a) To win the deal class must attains 74 or more. And 14 exams mean score is 73.5. Let your score in exam is x. Let us fine out the total score of 14 exams. So total score of 14 exams / 14 = 73.5 total score of 14 exams = 1029 1029 + x / 15 = 74 1029 + x = 74 * 15 x = 81 So must get 81 or more in order for the class to win the deal. (b) Not have to do the community service work class must attains 72. And 14 exams mean score is 73.5. Let your score in exam is x. Let us fine out the total score of 14 exams. So total score of 14 exams / 14 = 73.5 total score of 14 exams = 1029 1029 + x / 15 = 72 x= 72 * 15 -1029 x = 51 So must get at least 51 for the class not have to do the community service work.

Priya - Statistics HW 2.3

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