Find the total frequency.

Here's the solution for question 3: a) Calculate the sector angle for the speed 60 < s 70. Step 1: Find the total frequency. Total frequency is the sum of all frequencies: 3 + 14 + 6 + 12 + 15 = 50 Step 2: Find the frequency for the speed 60 < s 70. The frequency for 60 < s 64 is 6. The frequency for 64 < s 70 is 12. So, the frequency for 60 < s 70 is 6 + 12 = 18. Step 3: Calculate the sector angle. The sector angle is calculated as frequency for the grouptotal frequency × 360^. (18)/(50) × 360^ = 0.36 × 360^ = 129.6^ The sector angle for the speed 60 < s 70 is 129.6^. b) Estimate the number of cars whose speed was less than 68 km/h. Step 1: Sum the frequencies for speeds less than or equal to 64 km/h. Frequencies for 40 < s 50, 50 < s 60, and 60 < s 64 are 3, 14, and 6 respectively. Sum = 3 + 14 + 6 = 23. Step 2: Estimate the frequency for speeds between 64 km/h and 68 km/h. The interval 64 < s 70 has a frequency of 12 cars over a range of 70 - 64 = 6 km/h. We need to estimate for the range 64 < s 68, which is 68 - 64 = 4 km/h. Estimated frequency for 64 < s 68 = (4)/(6) × 12 = 8. Step 3: Add the frequencies. Total estimated number of cars with speed less than 68 km/h = 23 + 8 = 31. The estimated number of cars whose speed was less than 68 km/h is 31. c) Two cars are chosen at random. Calculate the probability that one car had a speed greater than 70 km/h and the other had the speed of 60 km/h or less. Step 1: Find the number of cars with speed greater than 70 km/h. This corresponds to the interval 70 < s 80, which has a frequency of 15. Step 2: Find the number of cars with speed 60 km/h or less. This corresponds to the intervals 40 < s 50 and 50 < s 60. Frequency = 3 + 14 = 17. Step 3: Calculate the total number of cars. Total frequency = 50. Step 4: Calculate the probability. Let A be the event that a car has speed > 70 km/h. P(A) = (15)/(50). Let B be the event that a car has speed 60 km/h. P(B) = (17)/(50). When two cars are chosen at random, the probability that one car had speed > 70 km/h AND the other had speed 60 km/h is given by: P(one A and one B) = P(A then B) + P(B then A) Since cars are chosen without replacement: P(A then B) = (15)/(50) × (17)/(49) P(B then A) = (17)/(50) × (15)/(49) So, the total probability is: 2 × (15)/(50) × (17)/(49) = 2 × (3)/(10) × (17)/(49) = (6)/(10) × (17)/(49) = (3)/(5) × (17)/(49) = (51)/(245) The probability is (51)/(245). d) Calculate an estimate of the mean speed. Step 1: Find the midpoint for each speed interval. • 40 < s 50: Midpoint = (40+50)/(2) = 45 • 50 < s 60: Midpoint = (50+60)/(2) = 55 • 60 < s 64: Midpoint = (60+64)/(2) = 62 • 64 < s 70: Midpoint = (64+70)/(2) = 67 • 70 < s 80: Midpoint = (70+80)/(2) = 75 Step 2: Calculate the sum of (midpoint × frequency). Sum = (45 × 3) + (55 × 14) + (62 × 6) + (67 × 12) + (75 × 15) Sum = 135 + 770 + 372 + 804 + 1125 Sum = 3206 Step 3: Calculate the total frequency. Total frequency = 50 (from part a). Step 4: Estimate the mean speed. Mean speed = Sum of (midpoint × frequency)Total frequency (3206)/(50) = 64.12 The estimate of the mean speed is 64.12 km/h. Send me the next one 📸