Question

The accompanying data represent the homework scores for material for a random sample of students in a college algebra course.

Answer 1

Answer:

$\begin{array}{ccc}{Class} & {Frequency} & {Relative\ Frequency} &{30-39} & {1} & {0.025} & {40-49} & {1} & {0.025} & {50 - 59} & {2} & {0.050} & {60 - 69} & {5} & {0.125} & {70 - 79} & {13} & {0.325} & {80 - 89} & {10} & {0.250} & {90 - 99} & {8} & {0.200} &{Total} & {40} & {1}\ \end{array}$

$P(x < 70) = 0.225$

Step-by-step explanation:

Given

$Lower = 30$

$Width = 10$

Solving (a): The relative frequency table

First, we construct the frequency table using the given parameters.

$\begin{array}{cc}{Class} & {Frequency} &{30-39} & {1} & {40-49} & {1} & {50 - 59} & {2} & {60 - 69} & {5} & {70 - 79} & {13} & {80 - 89} & {10} & {90 - 99} & {8} & {Total} & {40}\ \end{array}$

The relative frequency (RF) is calculated as:

$RF = \frac{Frequency}{Total}$

Using the above formula to calculate the relative frequency, the relative frequency table is:

$\begin{array}{ccc}{Class} & {Frequency} & {Relative\ Frequency} &{30-39} & {1} & {0.025} & {40-49} & {1} & {0.025} & {50 - 59} & {2} & {0.050} & {60 - 69} & {5} & {0.125} & {70 - 79} & {13} & {0.325} & {80 - 89} & {10} & {0.250} & {90 - 99} & {8} & {0.200} &{Total} & {40} & {1}\ \end{array}$

Solving (b): $P(x < 70)$

To do this, we add up the relative frequencies of classes less than 70.

i.e.

$P(x < 70) = [30 - 39] + [40 - 49] + [50 - 59] + [60 - 69]$

So, we have:

$P(x < 70) = 0.025 + 0.025 + 0.050 + 0.125$

$P(x < 70) = 0.225$