course: probability and statistics\nconsider the data below, calculate the mean and standard for grouped…

Explanation:

Step1: Calculate the mid - points

Let the values be (x_i) and frequencies be (f_i). The mid - points are already given as (x = 0,1,2,3,5,6) with frequencies (f = 5,13,2,5,1,3) respectively.

Step2: Calculate the mean (\bar{x})

The formula for the mean of grouped data is (\bar{x}=\frac{\sum_{i = 1}^{n}x_if_i}{\sum_{i = 1}^{n}f_i}). First, calculate (\sum_{i = 1}^{n}x_if_i): [ \begin{align*} \sum_{i = 1}^{n}x_if_i&=(0\times5)+(1\times13)+(2\times2)+(3\times5)+(5\times1)+(6\times3)\ &=0 + 13+4 + 15+5+18\ &=55 \end{align*} ] (\sum_{i = 1}^{n}f_i=5 + 13+2+5+1+3=29). So, (\bar{x}=\frac{55}{29}\approx1.90).

Step3: Calculate ((x_i-\bar{x})^2f_i) for each (i)

For (x_1 = 0): ((0 - 1.90)^2\times5=( - 1.90)^2\times5 = 3.61\times5=18.05). For (x_2 = 1): ((1 - 1.90)^2\times13=( - 0.90)^2\times13 = 0.81\times13 = 10.53). For (x_3 = 2): ((2 - 1.90)^2\times2=(0.10)^2\times2 = 0.01\times2=0.02). For (x_4 = 3): ((3 - 1.90)^2\times5=(1.10)^2\times5 = 1.21\times5 = 6.05). For (x_5 = 5): ((5 - 1.90)^2\times1=(3.10)^2\times1 = 9.61\times1=9.61). For (x_6 = 6): ((6 - 1.90)^2\times3=(4.10)^2\times3 = 16.81\times3 = 50.43). (\sum_{i = 1}^{n}(x_i-\bar{x})^2f_i=18.05+10.53+0.02+6.05+9.61+50.43 = 94.69).

Step4: Calculate the standard deviation (s)

The formula for the standard deviation of grouped data is (s=\sqrt{\frac{\sum_{i = 1}^{n}(x_i-\bar{x})^2f_i}{\sum_{i = 1}^{n}f_i - 1}}). (s=\sqrt{\frac{94.69}{29 - 1}}=\sqrt{\frac{94.69}{28}}\approx\sqrt{3.38}\approx1.84).

Answer:

Mean is approximately (1.90) and standard deviation is approximately (1.84).