Question

in a study of the time it takes an untrained mouse to run a maze, a researcher recorded these data in seconds.
find the mean of the data set. round to one decimal place, if necessary.
the mean time for an untrained mouse to run a maze was seconds.

Explanation:

Step1: Find the mid - point of each class

For class \(2.1 - 2.7\), mid - point \(x_1=\frac{2.1 + 2.7}{2}=2.4\)
For class \(2.8 - 3.4\), mid - point \(x_2=\frac{2.8+3.4}{2}=3.1\)
For class \(3.5 - 4.1\), mid - point \(x_3=\frac{3.5 + 4.1}{2}=3.8\)
For class \(4.2 - 4.8\), mid - point \(x_4=\frac{4.2+4.8}{2}=4.5\)
For class \(4.9 - 5.5\), mid - point \(x_5=\frac{4.9 + 5.5}{2}=5.2\)
For class \(5.6 - 6.2\), mid - point \(x_6=\frac{5.6+6.2}{2}=5.9\)

Step2: Calculate the sum of \(f\times x\)

Let \(f\) be the frequency.
\(f_1\times x_1=6\times2.4 = 14.4\)
\(f_2\times x_2=4\times3.1=12.4\)
\(f_3\times x_3=5\times3.8 = 19\)
\(f_4\times x_4=13\times4.5=58.5\)
\(f_5\times x_5=7\times5.2 = 36.4\)
\(f_6\times x_6=10\times5.9=59\)
\(\sum(f\times x)=14.4 + 12.4+19+58.5+36.4+59=209.7\)

Step3: Calculate the total frequency \(\sum f\)

\(\sum f=6 + 4+5+13+7+10=45\)

Step4: Calculate the mean

Mean \(\bar{x}=\frac{\sum(f\times x)}{\sum f}=\frac{209.7}{45}=4.66\approx4.7\)

Answer:

\(4.7\)