6.3: Measures of Central Tendency

Solution

\[\frac{79+86+82+94}{4}=85.25. \nonumber \]

Typically we round means to one more decimal place than the original data had. In this case, we would round 85.25 to 85.3. In a traditional US system, Marci would expect a B in the class.

Interpretation: If we equally distribute all the score points into the four exams, each exam will account for approximately 85.25 points.

Solution

Adding these values, we get 634 total TDs. Dividing by 31, the number of data values, we get \(\frac{634}{31} = 20.4516\). It would be appropriate to round this to 20.5.

It would be most correct for us to report that “The mean number of touchdown passes thrown in the NFL in the 2000 season was 20.5 passes,” but it is not uncommon to see the more casual word “average” used in place of “mean.

Interpretation: If we equally distribute the number of touchdowns among the 31 teams, each team will get approximately 20.5 points.

Solution

Calculating the mean by hand could get tricky if we try to type in all 100 values:

\[\frac{15 + \cdots + 15 + 20 + \cdots+20 + 25 + \cdots + 25 + \cdots}{100} \nonumber \]

We could calculate this more easily by noticing that adding 15 to itself six times is the same as \(15 \cdot 6=90\). Using this simplification, we get

\[\frac{15 \cdot 6+20 \cdot 8+25 \cdot 11+30 \cdot 17+35 \cdot 19+40 \cdot 20+45 \cdot 12+50 \cdot 7}{100}=\frac{3390}{100}=33.9 \nonumber \]

The mean household income of our sample is 33.9 thousand dollars ($33,900).

Interpretation: If the total amount of money earned by all households is equally split in each of them, each household would get $33,900.

Solution

\[\frac{15 \cdot 6+20 \cdot 8+25 \cdot 11+30 \cdot 17+35 \cdot 19+40 \cdot 20+45 \cdot 12+50 \cdot 7+5000 \cdot 1}{101}=\frac{8390}{101}=83.069 \nonumber \]

While 83.1 thousand dollars ($83,069) is the correct mean household income, it no longer represents a “typical” value.

Solution

We would start by listing the data in order. Luckily, it was already in decreasing order, so we can work with it without needing to reorder it first.

Since there are 31 data values, an odd number, the median will be the middle number, the 16^th data value (\frac{31}{2} = 15.5\), round up to 16, leaving 15 values below and 15 above). The 16^th data value is 20, so the median number of touchdown passes in the 2000 season was 20 passes. Notice that for this data, the median is fairly close to the mean we calculated earlier, 20.5.

Interpretation: In the 2000 season, about half of the teams had more than 20 touchdown passes and about half had fewer than 20 passes.

Solution

We start by listing the data in order: 2 4 5 5 6 7 7 8 8 10.

Since there are 10 data values, an even number, there is no one middle number. So we find the mean of the two middle numbers, 6 and 7, and get \(\dfrac{6+7}{2} = 6.5\).

The median quiz score was 6.5.

Interpretation: About half of the quiz scores are below 6.5 and half are above 6.5.

Solution

Here we have 100 data values. If we didn’t already know that, we could find it by adding the frequencies. Since 100 is an even number, we need to find the mean of the middle two data values -- the 50^th and 51^st data values. To find these values, we start counting up from the bottom:

\[\begin{array}{ll} \text{There are 6 data values of \$15, so} & \text{Values 1 to } 6 \text{ are \$15 thousand } \\ \text{The next 8 data values are \$20, so } & \text{Values 7 to } (6+8)=14 \text{ are \$20 thousand} \\ \text{The next 11 data values are \$25, so} & \text{ Values 15 to } (14+11)=25 \text{ are \$25 thousand} \\ \text{The next 17 data values are \$30, so} & \text{Values 26 to } (25+17)=42 \text{ are \$30 thousand} \\ \text{The next 19 data values are \$35, so} & \text{Values 43 to } (42+19)=61 \text{ are \$35 thousand} \end{array} \nonumber \]

From this we can tell that values 50 and 51 will be $35 thousand, and the mean of these two values is $35 thousand. The median income in this neighborhood is $35 thousand.

Interpretation: About half of the neighborhood's households have an income of $35,000 or less and about half have more than or equal to $35,000.

Solution

The mode is $40,000.

Interpretation: The most frequent household income is $40,000.