Averages from Grouped Frequency Tables

The modal class is the class with the highest frequency.

Find the midpoint of 4 and 8 first.

With 21 children, the median is the 11th height; add up the frequencies until you pass 11.

Multiply each midpoint by its frequency, add, then divide by the total frequency.

Think about what the table does not tell you about each person.

The total frequency is the number of dogs.

Part (a)

1. Add the frequencies: 6+13+11+5=35.
2. There are 35 dogs.

Part (b)

1. With 35 dogs, the median is the \frac{35+1}{2}th weight.
2. \frac{36}{2}=18, so the median is the weight of the 18th dog.

The midpoint of a class is halfway between its two ends.

Part (a)

1. Halfway between 170 and 180 is \frac{170+180}{2}=175.

Part (b)

1. The frequency of 170<h\le180 is 12.
2. 175\times12=2100

Use the midpoint of each class for every call in that class, then add up.

1. The exact length of each call is not known, so use the midpoint of each class as a representative length. For calls lasting from 0 to 4 minutes, the midpoint is \frac{0+4}{2}=2 minutes.
2. Estimate that each of the 6 calls in this class lasted 2 minutes: 2\times6=12 minutes altogether. The other estimated class totals are 6\times9=54, 10\times7=70 and 14\times3=42 minutes.
3. Estimated total time for all 25 calls: 12+54+70+42=178 minutes.

Multiply each midpoint by its frequency, add the results, then divide by the total frequency.

1. The exact parcel weights are not known, so use each class midpoint as a representative weight. For the first class, the midpoint is \frac{0+2}{2}=1 kg.
2. Estimate that each of the 7 parcels in this class weighs 1 kg: 1\times7=7 kg altogether. The other estimated class totals are 3\times15=45, 5\times12=60 and 7\times6=42 kg.
3. Estimated total weight: 7+45+60+42=154 kg. There are 40 parcels, so estimated mean weight =154\div40=3.85 kg per parcel.

With 30 times, the median is halfway between the 15th and 16th times.

1. With 30 times, the median is halfway between the 15th and 16th times.
2. The first two classes hold 8+5=13 times.
3. The first three classes hold 13+6=19 times, so the 15th and 16th times are both in the third class.
4. Answer: 10<t\le15. (This is not the modal class, 15<t\le20.)

Use the midpoints, divide by the total frequency, then round.

1. The exact leaf lengths are not known, so use each class midpoint as a representative length. For the first class, the midpoint is \frac{0+4}{2}=2 cm.
2. For 6 leaves in this class, the estimated total length is 2\times6=12 cm. The other estimated class totals are 6\times13=78, 10\times17=170, 14\times10=140 and 18\times3=54 cm.
3. Estimated total length: 12+78+170+140+54=454 cm. There are 49 leaves, so estimated mean length =454\div49=9.2653\ldots cm.
4. To 3 significant figures, the estimated mean length is 9.27 cm.

The midpoint of 0<t\le1.5 is 0.75. Find the other midpoints the same way.

1. The exact phone times are not known, so use each class midpoint as a representative time. For the first class, the midpoint is \frac{0+1.5}{2}=0.75 hours.
2. For 5 students in this class, the estimated total is 0.75\times5=3.75 hours. The other estimated class totals are 2.25\times12=27, 3.75\times9=33.75 and 5.25\times4=21 hours.
3. Estimated total phone time: 3.75+27+33.75+21=85.5 hours. There are 30 students, so estimated mean time =85.5\div30=2.85 hours per student.

Find which class the 20th and 21st waiting times are in.

1. With 40 times, the median is halfway between the 20th and 21st times.
2. Running totals: 3, 3+5=8, 8+7=15, 15+14=29.
3. The 20th and 21st times are in the fourth class, 30<w\le40.
4. So Lena is wrong: the middle row of the table is not where the middle value is.

Use the midpoints 15, 45, 75 and 105.

Part (a)

1. The exact homework times are not known, so use each class midpoint as a representative time. For the first class, the midpoint is \frac{0+30}{2}=15 minutes.
2. For 4 students in this class, the estimated total is 15\times4=60 minutes. The other estimated class totals are 45\times9=405, 75\times12=900 and 105\times5=525 minutes.
3. Estimated total homework time: 60+405+900+525=1890 minutes. There are 30 students, so estimated mean time =1890\div30=63 minutes per student.

Part (b)

1. Year 11’s estimated mean is 63 minutes, which is more than Year 10’s 52 minutes.
2. Year 11’s modal class is 60<t\le90, which is also higher than Year 10’s modal class, 30<t\le60.
3. Both comparisons support the statement that Year 11 students spent longer on homework on average.

The classes are not all the same width, so work out each midpoint carefully.

1. The exact amounts spent are not known, so use each class midpoint as a representative amount. For the first class, the midpoint is \frac{0+10}{2}=5 pounds.
2. For 12 customers in this class, the estimated total is 5\times12=£60. The other estimated class totals are 15\times18=£270, 30\times15=£450 and 60\times5=£300.
3. Estimated total spent: 60+270+450+300=£1080. There are 50 customers, so estimated mean spend =1080\div50=£21.60 to the nearest penny.

Read each bar carefully. For the group 0 to 4, the middle value is 2.

Part (a)

1. Total number of people: 14+18+11+7=50.
2. Fewer than 5 times means the group 0 to 4: 14 people.
3. \frac{14}{50}=\frac{7}{25}

Part (b)

1. The exact number of swims for each person is not known because the chart groups the answers. Use each group’s middle value as a representative value. Since the counts are whole numbers, the middle values are 2,\ 7,\ 12 and 17. For the first group, \frac{0+4}{2}=2.
2. For the 14 people in the 0 to 4 group, estimate 2 swims each: 2\times14=28 swims. The other estimated group totals are 7\times18=126, 12\times11=132 and 17\times7=119 swims.
3. Estimated total number of swims: 28+126+132+119=405.
4. There are 50 people. The estimated mean is the estimated total number of swims divided by the number of people: 405\div50=8.1 swims per person.

Add the two frequencies in each row to get one table for all 50 people.

1. First combine the two groups’ frequencies to count all 50 people: 2+4=6, 7+9=16, 11+8=19 and 5+4=9.
2. The exact times are not known, so use each class midpoint as a representative time. For the first class, the midpoint is \frac{0+10}{2}=5 minutes.
3. For 6 people in this class, the estimated total is 5\times6=30 minutes. The other estimated class totals are 15\times16=240, 25\times19=475 and 35\times9=315 minutes.
4. Estimated total time: 30+240+475+315=1060 minutes. There are 50 people, so estimated mean time =1060\div50=21.2 minutes per person.

The last class is wider than the others, so work out its midpoint carefully.

Part (a)

1. The exact bike prices are not known, so use each class midpoint as a representative price. For the first class, the midpoint is \frac{0+100}{2}=£50.
2. For 7 bikes in this class, the estimated total price is 50\times7=£350. The other estimated class totals are 150\times10=£1500, 250\times5=£1250 and 650\times3=£1950.
3. Estimated total price: 350+1500+1250+1950=£5050. There are 25 bikes, so estimated mean price =5050\div25=£202.

Part (b)

1. 17 of the 25 bikes cost £200 or less, but 3 bikes are in the £300 to £1000 class.
2. Those few high prices pull the estimated mean up to £202, above the price of most bikes. So the mean is not a good measure of a typical bike price here.

No one queued for more than 24 minutes, so the mean cannot be 66 minutes.

Part (a)

1. Kai found the estimated total time correctly: 264 minutes. But 264\div4=66 divides by the 4 classes, not the 24 people. That gives a mean per class, not a mean queue time per person.

Part (b)

1. The exact queue times are not known, so use each class midpoint as a representative time. For the first class, the midpoint is \frac{0+6}{2}=3 minutes.
2. For 5 people in this class, the estimated total is 3\times5=15 minutes. The other estimated class totals are 9\times9=81, 15\times7=105 and 21\times3=63 minutes.
3. Estimated total queue time: 15+81+105+63=264 minutes. There are 24 people, so estimated mean time =264\div24=11 minutes per person.

Find the estimate first, then compare it with 6 minutes.

Part (a)

1. Exact waiting times within each group are not known, so use each midpoint as a representative time. For the first group, \frac{0+4}{2}=2 minutes.
2. Estimate that each of the 18 customers in this group waited 2 minutes: 2\times18=36 minutes of waiting altogether for those customers.
3. The other midpoints are 6, 10 and 14 minutes. Their estimated totals are 6\times22=132, 10\times14=140 and 14\times6=84 minutes.
4. Estimated total waiting time across all customers: 36+132+140+84=392 minutes. Divide this estimated total by the 60 customers: 392\div60=6.533\ldots minutes per customer, which rounds to 6.5 minutes to 1 decimal place.

Part (b)

1. The estimate for the mean, 6.5 minutes, is more than 6 minutes.
2. So yes, she should employ another worker.