Question

extension. going beyond the basics.

1. the mean weight of six schoolbags is 9 pounds. the weights of five schoolbags are 8 pounds, 10 pounds, 8.6 pounds, 8.8 pounds and 9.5 pounds. find the weight of the last schoolbag.
2. the table shows the number of tickets of each price sold for a show.

price of ticket: $1, $2, $5, $10, $20
number of tickets sold: 12, 14, 9, 7, 8
a. calculate the mean price.
b. if each ticket price is increased by $2, and the frequency distribution remains unchanged, describe the effect on the mean and find the median price of the tickets.

Explanation:

Step1: Find the total weight of six school - bags

The formula for the mean $\bar{x}=\frac{\sum_{i = 1}^{n}x_{i}}{n}$. Given $n = 6$ and $\bar{x}=9$, the total weight of six school - bags is $n\times\bar{x}=6\times9 = 54$ pounds.

Step2: Find the total weight of five school - bags

Add the weights of the five given school - bags: $8 + 10+8.6 + 8.8+9.5=44.9$ pounds.

Step3: Find the weight of the last school - bag

Subtract the total weight of five school - bags from the total weight of six school - bags. Let the weight of the last school - bag be $x$. Then $x=54 - 44.9 = 9.1$ pounds.

Step4: Calculate the mean price for question 12a

The formula for the weighted mean $\bar{x}=\frac{\sum_{i = 1}^{k}x_{i}f_{i}}{\sum_{i = 1}^{k}f_{i}}$, where $x_{i}$ is the price and $f_{i}$ is the frequency.
$\sum_{i = 1}^{k}x_{i}f_{i}=1\times12 + 2\times14+5\times9 + 10\times7+20\times8=12+28 + 45+70+160=315$.
$\sum_{i = 1}^{k}f_{i}=12 + 14+9+7+8=50$.
The mean price $\bar{x}=\frac{315}{50}=6.3$ dollars.

Step5: Analyze the effect on the mean for question 12b

If each ticket price is increased by $2$, the new mean is $\bar{y}=\bar{x}+2$. Since $\bar{x}=6.3$, the new mean is $6.3 + 2=8.3$ dollars.
To find the median, first arrange the data in ascending order of price with frequencies. The total number of data points $n = 50$ (an even number).
The cumulative frequencies are: for price $1$, $cf_1 = 12$; for price $2$, $cf_2=12 + 14=26$; for price $5$, $cf_3=26 + 9=35$; for price $10$, $cf_4=35 + 7=42$; for price $20$, $cf_5=42 + 8=50$.
The median is the average of the 25th and 26th ordered values. Since the 25th and 26th values are in the group with price $2$, after increasing each price by $2$, the median is $2 + 2=4$ dollars.

Answer:

1. The weight of the last school - bag is 9.1 pounds.

12a. The mean price is 6.3 dollars.
12b. The mean increases by 2 dollars (new mean is 8.3 dollars) and the median is 4 dollars.