Question

a family of bears is going to a movie. the following data points represent the ticket and candy price (in dollars) for each bear. answer 2 questions about the data set. 1. sort the data from least to greatest. 0 1.50 2.50 3 4 4 4 7 7.50 2. find the interquartile range (iqr) of the data set. dollars

Explanation:

1. Sort the data from least to greatest

Step1: Identify the data points

The data points are \( 0, 1.50, 2.50, 3, 4, 4, 4, 7, 7.50 \).

Step2: Arrange in ascending order

We sort them by comparing each value. The sorted data is \( 0, 1.50, 2.50, 3, 4, 4, 4, 7, 7.50 \).

Step1: Find the median (Q2)

The data set has \( n = 9 \) values. The median is the middle value. For \( n = 9 \), the median is the \( \frac{9 + 1}{2}= 5^{\text{th}} \) value. The sorted data is \( 0, 1.50, 2.50, 3, 4, 4, 4, 7, 7.50 \), so the median (Q2) is \( 4 \).

Step2: Find the lower quartile (Q1)

The lower half of the data is \( 0, 1.50, 2.50, 3 \) (values before the median). The median of this lower half (Q1) is the average of the \( 2^{\text{nd}} \) and \( 3^{\text{rd}} \) values (since \( n = 4 \) for the lower half, median is \( \frac{2^{\text{nd}} + 3^{\text{rd}}}{2} \)). So \( Q1=\frac{1.50 + 2.50}{2}=\frac{4}{2}= 2 \).

Step3: Find the upper quartile (Q3)

The upper half of the data is \( 4, 4, 7, 7.50 \) (values after the median). The median of this upper half (Q3) is the average of the \( 2^{\text{nd}} \) and \( 3^{\text{rd}} \) values (since \( n = 4 \) for the upper half, median is \( \frac{2^{\text{nd}} + 3^{\text{rd}}}{2} \)). So \( Q3=\frac{4 + 7}{2}=\frac{11}{2}= 5.5 \).

Step4: Calculate IQR

The interquartile range is \( IQR = Q3 - Q1 \). Substituting the values of Q3 and Q1, we get \( IQR=5.5 - 2 = 3.5 \).

Answer:

\( 0, 1.50, 2.50, 3, 4, 4, 4, 7, 7.50 \)

2. Find the interquartile range (IQR) of the data set