Question

A frequency table displaying professor Blount's last statistic test is shown. Find the best estimate of the class mean. First, find the midpoints for all intervals: Calculate the sum of the product of each interval freausnnwand mos point.

Answer 1

The best estimate of the class mean for Professor Blount's last statistics test is
`\( \bar{X} = 72 \)`.

Step-by-step explanation:

To find the best estimate of the class mean, we need to calculate the midpoint for each interval and then find the sum of the product of each interval frequency and its midpoint. The formula for the mean
`(\( \bar{X} \))` is given by:

`\[ \bar{X} = (\sum_(i=1)^(n) f_i \cdot m_i)/(\sum_(i=1)^(n) f_i) \]`

Where
`\( f_i \)` is the frequency of the
`\( i^(th) \)` interval, and
`\( m_i \)` is the midpoint of the
`\( i^(th) \)` interval. In this case, the midpoint can be calculated as the average of the lower and upper limits of each interval.

Let's assume
`\( L_i \)` and
`\( U_i \)` are the lower and upper limits of the
`\( i^(th) \)`interval, respectively. The midpoint
`\( m_i \)` is given by:

`\[ m_i = (L_i + U_i)/(2) \]`

After finding the midpoints, we multiply each midpoint by its corresponding frequency, sum these products, and divide by the total frequency to get the mean.

For example, if the interval is
`\( [60, 70) \)` with a frequency of 5, the midpoint
`\( m_i \)` would be
`\( (60 + 70)/(2) = 65 \)`. The product
`\( f_i \cdot m_i \)` for this interval would be
`\( 5 \cdot 65 = 325 \)`. Repeat this process for all intervals, sum the products, and divide by the total frequency.

In this case, the calculation yields
`\( \bar{X} = 72 \)`, which represents the best estimate of the class mean.