Question

5.1 basics of probability distributions. compute the mean and standard deviation of a discrete random variable.
the following table shows the number of people that have either 0, 1, 2 or 3 cars registered in their name, from a sample of 286 people.
a) complete the table.

x (the number of cars)	number of people
1	59
2	68

|3|

b) find the mean number of cars owned for this sample. give your answer to at least 2 decimal places.
c) find the standard deviation of cars owned for this sample. give your answer to at least 2 decimal places.
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Explanation:

Step1: Calculate the number of people with 3 cars

We know the total number of people is 286. Let the number of people with 3 cars be $n$. Then $n = 286-(87 + 59+68)=286 - 214=72$.

Step2: Calculate the mean $\mu$

The formula for the mean of a discrete - random variable is $\mu=\sum_{i}x_ip_i$. Here, $p_i=\frac{\text{Number of people with }x_i\text{ cars}}{\text{Total number of people}}$.
$p_0=\frac{87}{286}\approx0.3042$, $p_1=\frac{59}{286}\approx0.2063$, $p_2=\frac{68}{286}\approx0.2378$, $p_3=\frac{72}{286}\approx0.2518$.
$\mu=(0\times0.3042)+(1\times0.2063)+(2\times0.2378)+(3\times0.2518)=0 + 0.2063+0.4756 + 0.7554=1.4373\approx1.44$.

Step3: Calculate the variance $\sigma^{2}$

The formula for the variance is $\sigma^{2}=\sum_{i}(x_i-\mu)^2p_i$.
$(0 - 1.44)^2\times0.3042=( - 1.44)^2\times0.3042 = 2.0736\times0.3042\approx0.6308$.
$(1 - 1.44)^2\times0.2063=( - 0.44)^2\times0.2063 = 0.1936\times0.2063\approx0.0400$.
$(2 - 1.44)^2\times0.2378=(0.56)^2\times0.2378 = 0.3136\times0.2378\approx0.0746$.
$(3 - 1.44)^2\times0.2518=(1.56)^2\times0.2518 = 2.4336\times0.2518\approx0.6128$.
$\sigma^{2}=0.6308 + 0.0400+0.0746 + 0.6128=1.3582$.

Step4: Calculate the standard deviation $\sigma$

The standard deviation $\sigma=\sqrt{\sigma^{2}}=\sqrt{1.3582}\approx1.17$.

Answer:

a) 72
b) 1.44
c) 1.17