Respuesta :
Answer:
The mean and standard deviation of the combined distribution is 16 and 7.192 respectively.
Step-by-step explanation:
We have given that a distribution consists of three components with frequencies 200, 250, and 300 having means 25, 10, and 15 and standard deviations 3, 4, and 5 respectively.
And we have to find the mean and standard deviation of the combined distribution.
Firstly let us represent some symbols;
[tex]n_1[/tex] = 200 [tex]\bar X_1[/tex] = 25 [tex]\sigma_1[/tex] = 3
[tex]n_2[/tex] = 250 [tex]\bar X_2[/tex] = 10 [tex]\sigma_2[/tex] = 4
[tex]n_3[/tex] = 300 [tex]\bar X_3[/tex] = 15 [tex]\sigma_3[/tex] = 5
Here, [tex]\bar X_1, \bar X_2 , \bar X_3[/tex] represent the means and [tex]\sigma_1,\sigma_2,\sigma_3[/tex] represent the standard deviations.
Now, as we know that Mean of the combined distribution is given by;
[tex]\bar X = \frac{n_1 \times \bar X_1+n_2 \times \bar X_2+n_3 \times \bar X_3}{n_1+n_2+n_3}[/tex]
Putting the above values in the formula we get;
[tex]\bar X = \frac{200 \times 25+250 \times 10+300 \times 15}{200+250+300}[/tex]
[tex]\bar X = \frac{5000+2500 +4500}{750}[/tex]
[tex]\bar X = \frac{12000}{750}[/tex] = 16
Similarly, the formula for combined standard deviation is given by;
[tex]\sigma = \sqrt{\frac{n_1\sigma_1^{2} + n_1(\bar X_1-\bar X)^{2}+n_2\sigma_2^{2} + n_2(\bar X_2-\bar X)^{2}+n_3\sigma_3^{2} + n_3(\bar X_3-\bar X)^{2} }{n_1+n_2+n_3} }[/tex]
[tex]\sigma = \sqrt{\frac{(200 \times 3^{2}) + 200 \times (25-16)^{2}+(250 \times 4^{2}) + 250 \times (10-16)^{2}+(300 \times 5^{2}) + 300 \times (15-16)^{2} }{200+250+300} }[/tex]
[tex]\sigma = \sqrt{\frac{1800 + (200 \times 81)+4000 + (250 \times 36)+7500 +( 300 \times 1) }{750} }[/tex]
[tex]\sigma = \sqrt{\frac{1800 + 16200+4000 + 9000+7500 +300 }{750} }[/tex]
[tex]\sigma = \sqrt{\frac{38800 }{750} }[/tex] = 7.192
Hence, the mean and standard deviation of the combined distribution is 16 and 7.192 respectively.