Binomial distribution help pls?

On average it is found that the failure rate for germination of geranium seeds, sold in packets of 10, is 0.8 seeds per packet. Find



1. the variance of the number of seeds per packet that fail to germinate.



i started with



X ~ B ( 10 , 0.8) correct??



2. the probability, to 3 decimal places, that a packet chosen at random will contain more than one seed that fails to germinate.



Thanks
Binomial distribution help pls?
Let X be the number of seeds that will germinate. X has the binomial distribution with n = 10 trials and success probability p = 0.8



In general, if X has the binomial distribution with n trials and a success probability of p then

P[X = x] = n!/(x!(n-x)!) * p^x * (1-p)^(n-x)

for values of x = 0, 1, 2, ..., n

P[X = x] = 0 for any other value of x.



The probability mass function is derived by looking at the number of combination of x objects chosen from n objects and then a total of x success and n - x failures.

Or, in other words, the binomial is the sum of n independent and identically distributed Bernoulli trials.



X ~ Binomial( n , p )



the mean of the binomial distribution is n * p = 8

the variance of the binomial distribution is n * p * (1 - p) = 1.6

the standard deviation is the square root of the variance = √ ( n * p * (1 - p)) = 1.264911



The Probability Mass Function, PMF,

f(X) = P(X = x) is:



P( X = 0 ) = 1.024e-07

P( X = 1 ) = 4.096e-06

P( X = 2 ) = 7.3728e-05

P( X = 3 ) = 0.000786432

P( X = 4 ) = 0.005505024

P( X = 5 ) = 0.02642412

P( X = 6 ) = 0.08808038

P( X = 7 ) = 0.2013266

P( X = 8 ) = 0.3019899

P( X = 9 ) = 0.2684355

P( X = 10 ) = 0.1073742





The Cumulative Distribution Function, CDF,

F(X) = P(X ≤ x) is:



x

∑ P(X = t) =

t = 0



P( X ≤ 0 ) = 1.024e-07

P( X ≤ 1 ) = 4.1984e-06

P( X ≤ 2 ) = 7.79264e-05

P( X ≤ 3 ) = 0.0008643584

P( X ≤ 4 ) = 0.006369382

P( X ≤ 5 ) = 0.0327935

P( X ≤ 6 ) = 0.1208739

P( X ≤ 7 ) = 0.3222005

P( X ≤ 8 ) = 0.6241904

P( X ≤ 9 ) = 0.8926258

P( X ≤ 10 ) = 1





1 - F(X) is:



n

∑ P(X = t) =

t = x



P( X ≥ 0 ) = 1

P( X ≥ 1 ) = 0.9999999

P( X ≥ 2 ) = 0.9999958

P( X ≥ 3 ) = 0.999922

P( X ≥ 4 ) = 0.9991356

P( X ≥ 5 ) = 0.9936306

P( X ≥ 6 ) = 0.9672065

P( X ≥ 7 ) = 0.8791261

P( X ≥ 8 ) = 0.6777995

P( X ≥ 9 ) = 0.3758096

P( X ≥ 10 ) = 0.1073742
Reply:hi

pl. mail me fr solution.

thankyou
Reply:yeah thats correct