Question

lacy raises mice. she measures the mass of each mouse three weeks after its birth. lacy records the data in a table.
masses of three - week - old mice (grams)
7\t7\frac{3}{8}\t7\frac{5}{8}\t6\frac{3}{4}
7\frac{1}{4}\t7\frac{1}{2}\t6\frac{3}{4}\t7\frac{1}{4}
7\frac{3}{4}\t7\frac{1}{4}\t7\frac{3}{4}\t7\frac{3}{8}

1. create a line plot to represent lacys data.
2. what is the most frequent mass of the three - week - old mice?
3. what is the difference in mass between the mice with the greatest mass and those with the least mass?
4. lacy says that most of the mice have a mass greater than 7\frac{1}{2} grams. do you agree with lacy? why?

Explanation:

Step1: Count the frequency of each mass

• \(7\): \(1\) time
• \(7\frac{3}{8}\): \(2\) times
• \(7\frac{5}{8}\): \(1\) time
• \(6\frac{3}{4}\): \(2\) times
• \(7\frac{1}{4}\): \(3\) times
• \(7\frac{1}{2}\): \(1\) time
• \(7\frac{3}{4}\): \(1\) time

Step2: Find the most frequent mass

The mass \(7\frac{1}{4}\) grams occurs \(3\) times, which is more frequent than other masses.

Step3: Find the greatest and least mass

The greatest mass is \(7\frac{5}{8}\) grams (\(7.625\) grams) and the least mass is \(6\frac{3}{4}\) grams (\(6.75\) grams).

$$7\frac{5}{8}-6\frac{3}{4}=\frac{61}{8}-\frac{27}{4}=\frac{61 - 54}{8}=\frac{7}{8}$$

Step4: Check Lacy's statement

Total number of mice \(=1 + 2+1 + 2+3+1+1=11\)
Number of mice with mass greater than \(7\frac{1}{2}\) grams (\(7.5\) grams) is \(1\) (the one with \(7\frac{5}{8}\) grams). \(1<\frac{11}{2}\)

Answer:

1. (Create a line - plot with \(x -\) axis labeled with the mass values \(6\frac{3}{4},7,7\frac{1}{4},7\frac{3}{8},7\frac{5}{8},7\frac{1}{2},7\frac{3}{4}\) and mark dots above each value according to their frequencies as calculated in Step 1)
2. \(7\frac{1}{4}\) grams
3. \(\frac{7}{8}\) grams
4. No. Because only \(1\) out of \(11\) mice has a mass greater than \(7\frac{1}{2}\) grams and \(1<\frac{11}{2}\)