there are many cell phone case options on the market. this discretionary item comes in a variety of colors…
there are many cell phone case options on the market. this discretionary item comes in a variety of colors, materials, thicknesses, protection levels, and more. amit runs a small business that sells computer and phone accessories. he has kept the following inventory of cell phone case sales for similar model phones: $19.99, $8.99, $35.99, $49.99, $49.99, $39.99, $99.99, $79.99, $35.99, $19.99, $49.99, $29.99, $49.99, $54.99, $35.99, $49.99, $19.99, $8.99, $29.99, $35.99\na. construct a frequency distribution for the data.\nb. use the frequency distribution to determine the mean.\nc. use the frequency distribution to determine the median and the mode
Answer
Explanation:
Step1: Organize data into classes
First, we choose appropriate class - intervals. Let's use: $8.99 - 14.99$, $15.99 - 21.99$, $22.99 - 28.99$, $29.99 - 35.99$, $36.99 - 42.99$, $43.99 - 49.99$, $50.99 - 56.99$, $75.99 - 81.99$, $95.99 - 101.99$. Then count the number of data values in each class.
Step2: Calculate frequency for each class
For the class $8.99 - 14.99$, frequency $f_1 = 2$ (since $8.99$ appears 2 times). For $15.99 - 21.99$, $f_2=4$ (as $19.99$ appears 4 times). For $22.99 - 28.99$, $f_3 = 0$. For $29.99 - 35.99$, $f_4=5$ (as $35.99$ appears 5 times). For $36.99 - 42.99$, $f_5 = 1$ (for $39.99$). For $43.99 - 49.99$, $f_6=6$ (as $49.99$ appears 6 times). For $50.99 - 56.99$, $f_7 = 1$ (for $54.99$). For $75.99 - 81.99$, $f_8 = 1$ (for $79.99$). For $95.99 - 101.99$, $f_9 = 1$ (for $99.99$). The frequency - distribution table is:
| Class Interval | Frequency |
|---|---|
| $8.99 - 14.99$ | 2 |
| $15.99 - 21.99$ | 4 |
| $22.99 - 28.99$ | 0 |
| $29.99 - 35.99$ | 5 |
| $36.99 - 42.99$ | 1 |
| $43.99 - 49.99$ | 6 |
| $50.99 - 56.99$ | 1 |
| $75.99 - 81.99$ | 1 |
| $95.99 - 101.99$ | 1 |
Step3: Calculate mid - points for each class
For $8.99 - 14.99$, mid - point $x_1=\frac{8.99 + 14.99}{2}=11.99$. For $15.99 - 21.99$, $x_2=\frac{15.99+21.99}{2}=18.99$. For $22.99 - 28.99$, $x_3=\frac{22.99 + 28.99}{2}=25.99$. For $29.99 - 35.99$, $x_4=\frac{29.99+35.99}{2}=32.99$. For $36.99 - 42.99$, $x_5=\frac{36.99+42.99}{2}=39.99$. For $43.99 - 49.99$, $x_6=\frac{43.99+49.99}{2}=46.99$. For $50.99 - 56.99$, $x_7=\frac{50.99+56.99}{2}=53.99$. For $75.99 - 81.99$, $x_8=\frac{75.99+81.99}{2}=78.99$. For $95.99 - 101.99$, $x_9=\frac{95.99+101.99}{2}=98.99$.
Step4: Calculate the mean
The formula for the mean $\bar{x}=\frac{\sum_{i = 1}^{n}f_ix_i}{\sum_{i = 1}^{n}f_i}$. $\sum_{i = 1}^{n}f_ix_i=2\times11.99 + 4\times18.99+0\times25.99 + 5\times32.99+1\times39.99+6\times46.99+1\times53.99+1\times78.99+1\times98.99$ $=23.98+75.96 + 0+164.95+39.99+281.94+53.99+78.99+98.99$ $=819.79$. $\sum_{i = 1}^{n}f_i=2 + 4+0+5+1+6+1+1+1=21$. $\bar{x}=\frac{819.79}{21}\approx39.04$.
Step5: Calculate the median
Since $n = 21$ (odd), the median is the $\left(\frac{n + 1}{2}\right)$ - th value. $\frac{21+1}{2}=11$ - th value. We calculate the cumulative frequencies: $CF_1 = 2$, $CF_2=2 + 4=6$, $CF_3=6+0 = 6$, $CF_4=6 + 5=11$. So the median class is $29.99 - 35.99$. We use the formula for median $M = L+\frac{\frac{n}{2}-CF}{f}\times w$, where $L = 29.99$, $CF = 6$, $f = 5$, $w=6$. $M=29.99+\frac{\frac{21}{2}-6}{5}\times6=29.99+\frac{10.5 - 6}{5}\times6=29.99+\frac{4.5}{5}\times6=29.99 + 5.4=35.39$.
Step6: Determine the mode
The mode is the value with the highest frequency. The highest frequency is $6$ and the corresponding class is $43.99 - 49.99$. Since $49.99$ appears most frequently, the mode is $49.99$.
Answer:
A. Frequency - distribution table:
| Class Interval | Frequency |
|---|---|
| $8.99 - 14.99$ | 2 |
| $15.99 - 21.99$ | 4 |
| $22.99 - 28.99$ | 0 |
| $29.99 - 35.99$ | 5 |
| $36.99 - 42.99$ | 1 |
| $43.99 - 49.99$ | 6 |
| $50.99 - 56.99$ | 1 |
| $75.99 - 81.99$ | 1 |
| $95.99 - 101.99$ | 1 |
| B. Mean $\approx39.04$ | |
| C. Median $\approx35.39$, Mode $=49.99$ |