81 991
Assignments Done
98,5%
Successfully Done
In December 2019

# Answer to Question #99868 in Statistics and Probability for Josephat

Question #99868
. In a binomial distribution consisting of 5 independent trials, the probability of 1 and 2 successes are 0.4096 and 0.2048 respectively. Calculate the mean, variance and mode of the distribution.
1
2019-12-04T09:36:25-0500

Probability mass function for binomial distribution @$pmf = \binom{n}{k}p^k(1-p)^{n-k}@$

For @$n=5\ and\ k=1@$ we have:

@$\binom{5}{1}p^1(1-p)^{5-1}= 5p(1-p)^4=0.4096@$, .

For @$n=5\ and \ k=2@$ :

@$\binom{5}{2}p^2(1-p)^{5-2}= \frac {5\cdot4}{1\cdot2}p^2(1-p)^3=10p^2(1-p)^3=0.2048@$.

@$\frac{5p(1-p)^4}{10p^2(1-p)^3}=\frac{0.4096}{0.2048},\ \frac{1-p}{2p}=2, 1-p=4p,\ 5p=1,\ p=1/5 @$

Since @$np=5\cdot1/5=1@$ is an integer,

then the mean and mode coincide and equal @$n\cdot p=1@$.

Variance @$V=n\cdot p\cdot (1-p)=5\cdot 1/5\cdot (1-1/5)=4/5@$.

Answer: mean = mode =1, variance = 4/5.

Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!