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Question 16
The masses, mkg, of some parcels are shown below. 4 15 14 11 12 3 1 18 13 2 16 10 Jack constructs this grouped frequency table to record the masses. | ... show full transcript
Step 1
Answer
Jack's table is unsuitable because the frequency intervals he has chosen overlap. The ranges '0 ≤ m < 5' and '5 ≤ m < 10' are correctly defined, but higher ranges such as '10 ≤ m < 15' and '15 ≤ m ≤ 20' create ambiguity. For instance, a mass of 15 kg could fit into both the '10 ≤ m < 15' and '15 ≤ m ≤ 20' categories, leading to confusion in data recording.
Step 2
Answer
To find the total number of students between 45 and 50 minutes, we calculate the area of the bars in the histogram from 45 to 50 minutes. According to the histogram:
Using the formula for area:
Similarly, for the interval from 50 to 60 minutes, the frequency density is 5 with the same width:
Adding these gives us:
Thus, 70 students took between 45 and 50 minutes to complete the race.
Step 3
Answer
To calculate the mean time, we use the midpoints of each interval in conjunction with the frequency densities:
Now, we sum up all the products:
Next, we need the total number of students, calculated earlier as 70. Thus, we compute the mean:
ext{Mean} = rac{5815}{70} = 83.0714
Hence, the estimated mean time taken to complete the race is approximately 83.07 minutes.
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