Question

question
stopping distance | frequency
48.0 ft | 1
49.0 ft | 2
50.0 ft | 3
51.0 ft | 2
52.0 ft | 2
the stopping distance for 10 cars traveling at 55 miles per hour is shown in the table above. what is the mean (average) stopping distance for the 10 cars?
25.0 ft
49.8 ft
50.0 ft
50.2 ft

Explanation:

Step1: Calculate total distance

Multiply each stopping distance by its frequency and sum them up.
For 48.0 ft: \(48.0\times1 = 48.0\)
For 49.0 ft: \(49.0\times2 = 98.0\)
For 50.0 ft: \(50.0\times3 = 150.0\)
For 51.0 ft: \(51.0\times2 = 102.0\)
For 52.0 ft: \(52.0\times2 = 104.0\)
Total distance = \(48.0 + 98.0 + 150.0 + 102.0 + 104.0 = 502.0\)

Step2: Calculate mean

Mean = Total distance / Total number of cars. Total number of cars is \(1 + 2 + 3 + 2 + 2 = 10\).
Mean = \(\frac{502.0}{10} = 50.2\) (Wait, there might be a typo in the options? Wait, let's recheck the calculation. Wait, 481=48, 492=98, 503=150, 512=102, 52*2=104. Sum: 48+98=146, 146+150=296, 296+102=398, 398+104=502. 502/10=50.2. So the correct option should be 50.2 ft (assuming the last option is 50.2 ft, maybe a typo in the original problem's option display).

Answer:

50.2 ft (assuming the last option is D. 50.2 ft, as per calculation)