Question

Create a probability distribution table from the frequency table. Round probabilities to the nearest hundredth.

Answer 1

Explanation:

From the table, we have:

Cummulative frequency = 5 + 10 + 2 + 9 + 33 + 12 + 15 + 3 + 1 = 90

We are to create a probability table. This is shown below:

$\begin{gathered} P=\frac{Frequency}{Cummulative\text{ frequency}} \\ \\ P(0)=\frac{5}{90}=\frac{1}{18}=0.055\approx0.06 \\ P(0)=0.06 \\ \\ P(1)=\frac{10}{90}=\frac{1}{9}=0.111\approx0.11 \\ P(1)=0.11 \\ \\ P(2)=\frac{2}{90}=\frac{1}{45}=0.022\approx0.02 \\ P(2)=0.02 \\ \\ P(3)=\frac{9}{90}=\frac{1}{10}=0.10 \\ P(3)=0.10 \\ \\ P(4)=\frac{33}{90}=\frac{11}{30}=0.366\approx0.37 \\ P(4)=0.37 \end{gathered}$

We procced to complete the values contained in the table, we have:

$\begin{gathered} P(5)=\frac{12}{90}=\frac{2}{15}=0.133\approx0.13 \\ P(5)=0.13 \\ \\ P(6)=\frac{15}{90}=\frac{1}{6}=0.166\approx0.17 \\ P(6)=0.17 \\ \\ P(7)=\frac{3}{90}=\frac{1}{30}=0.033\approx0.03 \\ P(7)=0.03 \\ \\ P(8)=\frac{1}{90}=0.011\approx0.01 \\ P(8)=0.01 \end{gathered}$

We will proceed to input this into a table, we have:

We thus have:

a) P(x < 4) is:

$\begin{gathered} =0.10+0.02+0.11+0.06 \\ =0.29 \end{gathered}$

b) P(x = 6) is:

$=0.17$

c) P(x ≥ 5) is:

$\begin{gathered} =0.13+0.17+0.03+0.01 \\ =0.34 \end{gathered}$