Question

A sample of 2,500 people was asked how many cups of coffee they drink in the morning. You are given the following sample information.Cups of Coffee Frequency0 7001 9002 6003 300total 2500The variance of the number of cups of coffee is (to four decimal places):

Answer 1

Answer:

The variance of the number of cups of coffee is 0.96

Step-by-step explanation:

We are given the following in the question:

Cups of coffee,x:    0        1       2         3

   Frequency,f(x): 700    900    600    300

Formula:

$E(x) = \dfrac{\sum fx}{\sum f}\\\\\text{Variance} = \dfrac{\sum fx^2}{\sum f} - (E(x) )^2$

Putting values, we get,

$E(x) = \dfrac{0(700) + 1(900) + 2(600) + 3(300)}{2500} = \dfrac{3000}{2500} = 1.2\\\\\dfrac{\sum fx^2}{\sum f} = \dfrac{0(700) + 1(900) + 4(600) + 9(300)}{2500}=2.4$

Variance =

$2.4 - (1.2)^2 = 0.96$

Thus, the variance of the number of cups of coffee is 0.96

Answer 2

The variance on the number of cups of coffee is 0.9604

Data;

• Total number of people = 2500
• f1 = 700
• f2 = 900
• f3 = 600
• f4 = 300

Variance

To solve the variance, we have to find the mean of the data.

$mean(\mu) = \frac{\sum _f_*_x}{\sum _F}$

Let's substitute the values into the formula

$\mu = \frac{0*700 + 1 * 900 + 2 * 600 + 3 * 300 }{2500}\\\mu = \frac{3000}{2500} \\\mu = 1.2$

The formula of variance is given as;

$V = \frac{\sum _f_*_x_^2 - n\mu ^2}{n-1} \\V = \frac{0^2 * 700 + 1^2 * 900 + 2^2 * 600 + 3^2 * 300 - 2500 * 1.2^2}{2500-1} \\V = \frac{0+900+(4*60)+(9*300)-2500*1.44}{2499} \\V = 0.9604$

The variance on the number of cups of coffee is 0.9604

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