Word problems finding the mean

Here we will learn to solve the
three important types of word problems on arithmetic mean (average). The
questions are mainly based on average (arithmetic mean), weighted average and average
speed.

How to solve average (arithmetic mean) word problems?

To solve various problems we need to follow the uses of the formula for calculating average (arithmetic mean)

Average = (Sums of the observations)/(Number of observations)

Follow the explanation to solve the word problems on arithmetic mean (average):

1. The heights of five runners are 160 cm, 137 cm, 149 cm, 153 cm and 161 cm respectively. Find the mean height per runner.

Solution:

Mean height = Sum of the heights
of the runners/number of runners

= (160 + 137 + 149 + 153 + 161)/5 cm

= 760/5 cm

= 152 cm.

Hence, the mean height is 152
cm.

2. Find
the mean of the first five prime numbers.

Solution:

The first five prime numbers are
2, 3, 5, 7 and 11.

Mean
= Sum of the first five prime numbers/number of prime numbers

= (2 + 3 + 5 + 7 + 11)/5

= 28/5

= 5.6

Hence,
their mean is 5.6

3. Find the mean of the
first six multiples of 4.

Solution:

The first six multiples of 4 are
4, 8, 12, 16, 20 and 24.

Mean = Sum of the first
six multiples of 4/number of multiples

= (4 + 8 + 12 + 16 + 20 + 24)/6

= 84/6

= 14.

Hence,
their mean is 14.

4. Find the arithmetic mean of the first 7 natural numbers.

Solution:

The first 7 natural numbers are 1,
2, 3, 4, 5, 6 and 7.

Let x denote their arithmetic mean.

Then mean = Sum of the first 7 natural numbers/number of natural numbers

x = (1 + 2 + 3 + 4 + 5 + 6 + 7)/7

= 28/7

= 4

Hence, their mean is 4.

5. If the mean of 9, 8, 10, x, 12 is 15, find the value of x.

Solution:

Mean of the given numbers = (9 + 8 + 10 + x + 12)/5 = (39 + x)/5

According to the problem, mean = 15 (given).

Therefore, (39 + x)/5 = 15

⇒ 39 + x = 15 × 5

⇒ 39 + x = 75

⇒ 39 — 39 + x = 75 — 39

⇒ x = 36

Hence, x = 36.

More examples on the worked-out word problems
on
arithmetic mean:

6. If
the mean of five observations x, x + 4, x + 6, x + 8 and x + 12 is 16,
find the value of x.

Solution:
Mean of the
given observations


= x + (x + 4) + (x + 6) + (x + 8) + (x + 12)/5


= (5x + 30)/5

According to the problem, mean =
16 (given).

Therefore, (5x + 30)/5 = 16

⇒ 5x + 30 = 16 × 5

⇒ 5x + 30 = 80

⇒ 5x + 30 — 30 = 80 — 30

⇒ 5x = 50

⇒ x = 50/5

⇒ x = 10

Hence, x = 10.

148 + 153 + 146 + 147 + 154

7. The mean of 40 numbers was found to be 38.
Later on, it was detected that
a number 56 was misread as 36. Find
the correct mean of given numbers.

Solution:

Calculated mean of 40 numbers =
38.

Therefore, calculated sum of these numbers = (38 × 40) = 1520.

Correct sum of these numbers

= [1520 — (wrong item) + (correct item)]

= (1520 — 36 + 56)

= 1540.

Therefore,
the correct mean = 1540/40 = 38.5.

8. The mean of the heights of 6 boys is 152
cm. If the individual heights of five
of them are 151 cm, 153 cm, 155 cm, 149 cm and 154 cm, find the
height of the sixth boy.

Solution:

Mean height of 6 boys = 152 cm.

Sum of the heights of 6 boys =
(152 × 6) = 912 cm

Sum of the heights of 5 boys =
(151 + 153 + 155 + 149 + 154) cm = 762
cm.

Height of the sixth boy

= (sum of the heights of 6 boys) — (sum of the heights of 5 boys)

= (912 — 762) cm = 150 cm.


Hence,
the height of the sixth girl is 150 cm.

Statistics

Arithmetic Mean

Word Problems on Arithmetic Mean

Properties of Arithmetic Mean

Problems Based on Average

Properties Questions on Arithmetic Mean

9th Grade Math

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In our guide on descriptive statistics, we’ve shown you the fundamental concepts involved in the descriptive branch of statistics. From measures of central tendency to visualizing a standard normal distribution, we’ve taught you how to calculate measures and challenged you with practice problems.

In this section, you’ll find statistical word problems that require you to apply the knowledge you’ve learned with us thus far. Good luck!

Practice Problems

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Let’s go

Problem 1: Central Tendency

You have gathered data on customers who come into your store for a week, including what they have bought and how many items they have bought. You’re interested in learning whether it’s worth it for you or not to start a loyalty program and, if so, on what products. Given the table below, use a measure of central tendency to solve your problem.

Customer Apples Hummus Bread Oil
A 3 15 1 0
B 5 6 2 1
C 1 9 2 0

Problem 2: Interpreting Variability

You’re studying the effects of a weed killer on a specific weed type. Given the following information, what can you say about the reliability of the weed killer on reducing the length of the weeds studied using measures of variability?

Original Length (inches) Length After 1 Month of Treatment (inches)
1 0.5
2 1.9
1.5 0.2
2.3 1.7

Problem 3: Finding the Group Mean

Given the following information about 2017 estimated voter turnout, what can you say about young voters between 18 and 40 years of age? Data taken from YouGov UK.

Age Group People Sampled Percentage Who Voted
18-19 1077 57
20-24 2679 59
25-29 3648 64
30-39 7912 61
40-49 8237 66
50-59 10718 71
60-69 12388 77
70+ 5956 84

Problem 4: Coefficient of Variation

You have the following information about estimated voter turnout. Which data set will be better to use in your research paper if you’re looking for the most accurate data?

Data Set 1 2 3 4

    [ bar{x} ]

42 100 3 500 9 600 17 500

    [ s ]

1 000 150 210 3 000

    [ n ]

52 300 4 000 10 500 23 000

Problem 5: Quartiles

Interpret the characteristics of the following chart.

Skewed Boxplot 2

Problem 6: Changing Units

You are given the following information about hair growth caused by hair supplements in one month. Find the new mean and variance for each supplement after 1 month.

Supplement Mean Hair Length Before Variance Before Hair Growth After 1 Month
A 30.6 0.3 103%
B 38.2 0.1 105%
C 32.4 0.5 101%
D 34.5 0.2 104%

Solutions to Practice Problems

In the previous section, you were asked to solve word problems covering the basics of descriptive statistics. Keep in mind that in statistics, there is rarely only one right answer.

Solution Problem 1

In this task, you were asked to:

  • Find the best measure of central tendency for the question
  • Use its calculation to respond whether or not it’s worth it to start a loyalty program
  • If it is worth it, on which products

In this question, since we want to know the most frequent customers and what they buy the most of, we use the mode.

Customer Apples Hummus Bread Oil Total
A 3 15 1 0 19
B 5 6 2 1 14
C 1 9 2 0 12
Total 9 30 5 1

Each customer buys over 10 products each week, which can indicate that a loyalty program will be beneficial. The product they should put on the loyalty program, or the mode of the products, is hummus.

Solution Problem 2

In this task we were asked to make a statement about the reliability of a weed killer. To do this, we follow the steps below.

Original Length (inches) Length After 1 Month of Treatment (inches) Original Length — Length After Treatment
1 0.5 0.5
2 1.9 0.1
1.5 0.2 1.3
2.3 1.7 0.6
Variance 0.25
Mean 0.63

Looking at the variance and mean of the differences in length before and after treatment, we can see that the variance is almost half the mean. Meaning, the values are spread quite far around the mean, which indicates a lower reliability.

Solution Problem 3

Given the information in the table, we were asked to make a statement about young voters. To do this, we can find the group mean of people who voted aged between 18 and 40 by following the steps below.

Age Group People Sampled Percentage Who Voted f_{i} x_{m} f_{i}*x_{m}
n_{i} f_{i}% n_{i} * f_{i}%/100 Average of upper and lower group limits Frequency * midpoint
18-19 1077 57 614 18.5 11357
20-24 2679 59 1581 22 34773
25-29 3648 64 2335 27 63037
30-39 7912 61 4826 34.5 166508
Total 9356 275676

The group mean is found by,

    [ x_{group} = frac{Sigma(f_{i}*x_{m})}{n} ]

    [ x_{group} = frac{275676}{9356} ]

    [ x_{group} = 29.5 ]

Which tells us that on average, the young people who actually voted were aged between 25 to 39.

Solution Problem 4

In this problem, you were asked to

  • Find the most accurate data between the 4 data sets

Since we want to understand the variability between data sets, we use the coefficient of variation, found below.

Data Set

1 2 3

4

CV

    [ dfrac{1 000}{42 100} ]

    [ *100% ]

    [ = 2.4% ]

    [ dfrac{150}{3 500} ]

    [ *100% ]

    [ = 4.3% ]

    [ dfrac{210}{9 600} ]

    [ *100% ]

    [ = 2.2% ]

    [ dfrac{3 000}{17 500} ]

    [ *100% ]

    [ = 17.1% ]

For the third data set, the standard deviation is only 2.2% of the mean, meaning it has the least variability and will be more accurate.

Solution Problem 5

In this problem, you were tasked with interpreting the following chart.

Skewed Boxplot 2

We can interpret general characteristics about the distribution,

  • Centre: The median is located between 35 and 40
  • Spread: The data appears to be unevenly spread around the median
  • Skew: The data are skewed to the right. Quartile 1 is closer to the median, while quartile 3 is farther, indicating that lesser values are located closer together than higher values.

Solution Problem 6

This problem involved:

  • Calculating the new mean and variance for each supplement after one months of growth

In order to do this, we must use the rules for changing units.

Supplement Mean Hair Length Before Variance Before Hair Growth After 1 Month New Mean New Variance
A 30.6 0.3 0.103

    [ = 30.6 ]

    [ *0.103 ]

    [ = 31.5 ]

    [ = 0.3 ]

    [ *(0.103^2) ]

    [ = 0.32 ]

B 38.2 0.1 0.105

    [ = 38.2 ]

    [ *0.105 ]

    [ = 40.1 ]

    [ = 0.1 ]

    [ *(0.105^2) ]

    [ = 0.11 ]

C 32.4 0.5 0.101

    [ = 32.4 ]

    [ *0.101 ]

    [ = 32.7 ]

    [ = 0.5 ]

    [ *(0.101^2) ]

    [ = 0.51 ]

D 34.5 0.2 0.104

    [ = 34.5 ]

    [ *0.104 ]

    [ = 35.9 ]

    [ = 0.2 ]

    [ *(0.104^2) ]

    [ = 0.22 ]

Objective: I know how to solve word problems that involve the mean.

In statistics, mode, median and mean are typical values to represent a pool of numerical observations. They are calculated from the pool of observations.

Modeis the most common value among the given observations. For example, a person who sells ice creams might want to know which flavor is the most popular.

Median is the middle value, dividing the number of data into 2 halves. In other words, 50% of the observations is below the median and 50% of the observations is above the median.

Mean is the average of all the values. For example, a teacher may want to know the average marks of a test in his class.

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the arithmetic mean (simply the mean or average) is the sum of a collection of numbers divided by the count of numbers in the collection. The arithmetic mean is the most commonly used and readily understood measure of central tendency in a data set. It is a statistical characteristic of the center of a given data set.

Arithmetic mean is a quantity very sensitive to extreme values, so the median and mode are also used in practice. Mode is the most common value. The median halves the ordered file.

Number of problems found: 546

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