Question 871914

<table border=1>
  <tr>
    <th>Number of Bedrooms</th>
    <th>Frequency</th>
  </tr>
  <tr>
    <td>2</td>
    <td>12</td>
  </tr>
  <tr>
    <td>3</td>
    <td>40</td>
  </tr>
  <tr>
    <td>4</td>
    <td>70</td>
  </tr>
  <tr>
    <td>5</td>
    <td>3</td>
  </tr>
</table>


There are 12+40+70+3 = 125 homes being surveyed.


"3 or fewer" means "at most 3" (think of it as the ceiling or as high as you can go). X = 2 is the lowest you can go on the table while X = 3 is the highest we want. So we just need to calculate P(X = 2) and P(X = 3) then add up them. The variable X is defined as X = number of bedrooms.



From the table, we get the following probabilities


P(X = 2) = 12/125
P(X = 3) = 40/125


Use them to calculate P(X <= 3)


P(X <= 3) = P(X = 2) + P(X = 3)
P(X <= 3) = 12/125 + 40/125
P(X <= 3) = (12 + 40)/125
P(X <= 3) = 52/125
P(X <= 3) = 0.416



So the final answer is <font color="red">0.416</font>