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Asked on July 17, 2026English

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Accepted Answer · 2026-07-17T05:11:11.934Z

Here is the solution to question 7: a) To calculate the mean, median, and standard deviation, we first construct a frequency distribution table with midpoints, deviations, and cumulative frequencies. Given assumed mean A = 94.5. Class width c = 10. | Length (cm) | Class Boundaries | Midpoint (x) | Frequency (f) | d = x - A | fd | d^2 | fd^2 | Cumulative Frequency (cf) | | :---------- | :--------------- | :------------- | :-------------- | :---------- | :--- | :---- | :----- | :------------------------ | | 50-59 | 49.5-59.5 | 54.5 | 10 | -40 | -400 | 1600 | 16000 | 10 | | 60-69 | 59.5-69.5 | 64.5 | 20 | -30 | -600 | 900 | 18000 | 30 | | 70-79 | 69.5-79.5 | 74.5 | 45 | -20 | -900 | 400 | 18000 | 75 | | 80-89 | 79.5-89.5 | 84.5 | 52 | -10 | -520 | 100 | 5200 | 127 | | 90-99 | 89.5-99.5 | 94.5 | 43 | 0 | 0 | 0 | 0 | 170 | | 100-109 | 99.5-109.5 | 104.5 | 18 | 10 | 180 | 100 | 1800 | 188 | | 110-119 | 109.5-119.5 | 114.5 | 12 | 20 | 240 | 400 | 4800 | 200 | | Total | | | f = 200 | | fd = -2000 | | fd^2 = 63800 | | i) mean length; Step 1: Calculate the mean using the assumed mean formula. The formula for the mean (x) using an assumed mean (A) is: x = A + ( fd)/( f) Substitute the values from the table: x = 94.5 + (-2000)/(200) x = 94.5 - 10 x = 84.5 cm The mean length is 84.5 cm. ii) median; Step 1: Determine the median position and median class. Total frequency N = f = 200. The median position is (N)/(2) = (200)/(2) = 100^th value. From the cumulative frequency column, the 100^th value falls in the class 80-89 (cf = 127). So, the median class is 80-89. Step 2: Apply the median formula for grouped data. Lower boundary of median class (L) = 79.5 Frequency of median class (f_m) = 52 Cumulative frequency of the class before the median class (cf_b) = 75 Class width (c) = 10 The formula for the median is: Median = L + ((N/2 - cf_b)/(f_m))c Substitute the values: Median = 79.5 + ((100 - 75)/(52))10 Median = 79.5 + ((25)/(52))10 Median = 79.5 + 4.80769 Median ≈ 84.31 cm The median is 84.31 cm. iii) standard deviation. Step 1: Apply the standard deviation formula using the assumed mean. The formula for standard deviation () using an assumed mean is: = sqrt(( fd^2)/( f) - (( fd)/( f))^2) Substitute the values from the table: = sqrt((63800)/(200) - ((-2000)/(200))^2) = sqrt(319 - (-10)^2) = sqrt(319 - 100) = sqrt(219) ≈ 14.80 cm The standard deviation is 14.80 cm. b) Table 2 shows the diameters of 100 metal pipes. Total number of pipes N = 100. Mean diameter x = 41.95. First, we determine the midpoints (x) for each class. Class width c = 3. | Diameter (cm) | Class Boundaries | Midpoint (x) | Frequency (f) | fx | | :------------ | :--------------- | :------------- | :-------------- | :--- | | 33-35 | 32.5-35.5 | 34 | 7 | 238 | | 36-38 | 35.5-38.5 | 37 | a | 37a | | 39-41 | 38.5-41.5 | 40 | 22 | 880 | | 42-44 | 41.5-44.5 | 43 | 36 | 1548 | | 45-47 | 44.5-47.5 | 46 | b | 46b | | 48-50 | 47.5-50.5 | 49 | 9 | 441 | | Total | | | f = 100 | fx = 3107 + 37a + 46b | i) Given that the mean is 41.95, determine the value of a and b; Step 1: Formulate equations based on total frequency and mean. From the total frequency: 7 + a + 22 + 36 + b + 9 = 100 74 + a + b = 100 a + b = 26 (Equation 1) From the mean formula x = ( fx)/( f): 41.95 = (3107 + 37a + 46b)/(100) 4195 = 3107 + 37a + 46b 37a + 46b = 4195 - 3107 37a + 46b = 1088 (Equation 2) Step 2: Solve the system of linear equations for a and b. From Equation 1, express a in terms of b: a = 26 - b. Substitute this into Equation 2: 37(26 - b) + 46b = 1088 962 - 37b + 46b = 1088 9b = 1088 - 962 9b = 126 b = (126)/(9) b = 14 Now substitute b=14 back into a = 26 - b: a = 26 - 14 a = 12 The values are a = 12 and b = 14. ii) hence calculate the mode. Step 1: Identify the modal class. The frequencies are 7, a=12, 22, 36, b=14, 9. The highest frequency is 36, which corresponds to the class 42-44. So, the modal class is 42-44. Step 2: Apply the mode formula for grouped data. Lower boundary of modal class (L_m) = 41.5 Frequency of modal class (f_m) = 36 Frequency of the class before the modal class (f_1) = 22 Frequency of the class after the modal class (f_2) = 14 Class width (c) = 3 The formula for the mode is: Mode = L_m + ((f_m - f_1)/((f_m - f_1) + (f_m - f_2)))c Substitute the values: Mode = 41.5 + ((36 - 22)/((36 - 22) + (36 - 14)))3 Mode = 41.5 + ((14)/(14 + 22))3 Mode = 41.5 + ((14)/(36))3 Mode = 41.5 + (14)/(12) Mode = 41.5 + 1.1666... Mode ≈ 42.67 cm The mode is 42.67 cm. 3 done, 2 left today. You're making progress.

Accepted Answer · 2026-07-17T05:11:11.934Z

Here is the solution to question 7: a) To calculate the mean, median, and standard deviation, we first construct a frequency distribution table with midpoints, deviations, and cumulative frequencies. Given assumed mean A = 94.5. Class width c = 10. | Length (cm) | Class Boundaries | Midpoint (x) | Frequency (f) | d = x - A | fd | d^2 | fd^2 | Cumulative Frequency (cf) | | :---------- | :--------------- | :------------- | :-------------- | :---------- | :--- | :---- | :----- | :------------------------ | | 50-59 | 49.5-59.5 | 54.5 | 10 | -40 | -400 | 1600 | 16000 | 10 | | 60-69 | 59.5-69.5 | 64.5 | 20 | -30 | -600 | 900 | 18000 | 30 | | 70-79 | 69.5-79.5 | 74.5 | 45 | -20 | -900 | 400 | 18000 | 75 | | 80-89 | 79.5-89.5 | 84.5 | 52 | -10 | -520 | 100 | 5200 | 127 | | 90-99 | 89.5-99.5 | 94.5 | 43 | 0 | 0 | 0 | 0 | 170 | | 100-109 | 99.5-109.5 | 104.5 | 18 | 10 | 180 | 100 | 1800 | 188 | | 110-119 | 109.5-119.5 | 114.5 | 12 | 20 | 240 | 400 | 4800 | 200 | | Total | | | f = 200 | | fd = -2000 | | fd^2 = 63800 | | i) mean length; Step 1: Calculate the mean using the assumed mean formula. The formula for the mean (x) using an assumed mean (A) is: x = A + ( fd)/( f) Substitute the values from the table: x = 94.5 + (-2000)/(200) x = 94.5 - 10 x = 84.5 cm The mean length is 84.5 cm. ii) median; Step 1: Determine the median position and median class. Total frequency N = f = 200. The median position is (N)/(2) = (200)/(2) = 100^th value. From the cumulative frequency column, the 100^th value falls in the class 80-89 (cf = 127). So, the median class is 80-89. Step 2: Apply the median formula for grouped data. Lower boundary of median class (L) = 79.5 Frequency of median class (f_m) = 52 Cumulative frequency of the class before the median class (cf_b) = 75 Class width (c) = 10 The formula for the median is: Median = L + ((N/2 - cf_b)/(f_m))c Substitute the values: Median = 79.5 + ((100 - 75)/(52))10 Median = 79.5 + ((25)/(52))10 Median = 79.5 + 4.80769 Median ≈ 84.31 cm The median is 84.31 cm. iii) standard deviation. Step 1: Apply the standard deviation formula using the assumed mean. The formula for standard deviation () using an assumed mean is: = sqrt(( fd^2)/( f) - (( fd)/( f))^2) Substitute the values from the table: = sqrt((63800)/(200) - ((-2000)/(200))^2) = sqrt(319 - (-10)^2) = sqrt(319 - 100) = sqrt(219) ≈ 14.80 cm The standard deviation is 14.80 cm. b) Table 2 shows the diameters of 100 metal pipes. Total number of pipes N = 100. Mean diameter x = 41.95. First, we determine the midpoints (x) for each class. Class width c = 3. | Diameter (cm) | Class Boundaries | Midpoint (x) | Frequency (f) | fx | | :------------ | :--------------- | :------------- | :-------------- | :--- | | 33-35 | 32.5-35.5 | 34 | 7 | 238 | | 36-38 | 35.5-38.5 | 37 | a | 37a | | 39-41 | 38.5-41.5 | 40 | 22 | 880 | | 42-44 | 41.5-44.5 | 43 | 36 | 1548 | | 45-47 | 44.5-47.5 | 46 | b | 46b | | 48-50 | 47.5-50.5 | 49 | 9 | 441 | | Total | | | f = 100 | fx = 3107 + 37a + 46b | i) Given that the mean is 41.95, determine the value of a and b; Step 1: Formulate equations based on total frequency and mean. From the total frequency: 7 + a + 22 + 36 + b + 9 = 100 74 + a + b = 100 a + b = 26 (Equation 1) From the mean formula x = ( fx)/( f): 41.95 = (3107 + 37a + 46b)/(100) 4195 = 3107 + 37a + 46b 37a + 46b = 4195 - 3107 37a + 46b = 1088 (Equation 2) Step 2: Solve the system of linear equations for a and b. From Equation 1, express a in terms of b: a = 26 - b. Substitute this into Equation 2: 37(26 - b) + 46b = 1088 962 - 37b + 46b = 1088 9b = 1088 - 962 9b = 126 b = (126)/(9) b = 14 Now substitute b=14 back into a = 26 - b: a = 26 - 14 a = 12 The values are a = 12 and b = 14. ii) hence calculate the mode. Step 1: Identify the modal class. The frequencies are 7, a=12, 22, 36, b=14, 9. The highest frequency is 36, which corresponds to the class 42-44. So, the modal class is 42-44. Step 2: Apply the mode formula for grouped data. Lower boundary of modal class (L_m) = 41.5 Frequency of modal class (f_m) = 36 Frequency of the class before the modal class (f_1) = 22 Frequency of the class after the modal class (f_2) = 14 Class width (c) = 3 The formula for the mode is: Mode = L_m + ((f_m - f_1)/((f_m - f_1) + (f_m - f_2)))c Substitute the values: Mode = 41.5 + ((36 - 22)/((36 - 22) + (36 - 14)))3 Mode = 41.5 + ((14)/(14 + 22))3 Mode = 41.5 + ((14)/(36))3 Mode = 41.5 + (14)/(12) Mode = 41.5 + 1.1666... Mode ≈ 42.67 cm The mode is 42.67 cm. 3 done, 2 left today. You're making progress.

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