Need some help with statistics question, please.?

I need some help with this problem please. Thank you!





A poll found that about 40% of employees missed work due to a back injury. Let x be the number of sampled workers who have missed work due to back injury.





a) Explain why x is approximately a binomial random variable.





b) Use the poll data to estimate p for the binomial random variable of part a.





c) A random sample of 10 workers is to be drawn from a particular manufacturing plant. Use the p from part b to find the mean and standard deviation of x.





d) From the sample in part c, find the probability that exactly one worker missed work due to back injury and the probability that more than one worker missed work due to a back injury.





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Need some help with statistics question, please.?
a) The binomial is the sum of n independent and identically distributed Bernoulli trails. each work in the sample is a Bernoulli trial, either the did or did not miss work due to a back injury.





b) p = 0.40





c and c)





Let X be the number of workers who missed work due to a back injury. X has the binomial distribution with n = 10 trials and success probability p = 0.4





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 = 4


the variance of the binomial distribution is n * p * (1 - p) = 2.4


the standard deviation is the square root of the variance = √ ( n * p * (1 - p)) = 1.549193





The Probability Mass Function, PMF,


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





P( X = 0 ) = 0.006046618


P( X = 1 ) = 0.04031078


P( X = 2 ) = 0.1209324


P( X = 3 ) = 0.2149908


P( X = 4 ) = 0.2508227


P( X = 5 ) = 0.2006581


P( X = 6 ) = 0.1114767


P( X = 7 ) = 0.04246733


P( X = 8 ) = 0.01061683


P( X = 9 ) = 0.001572864


P( X = 10 ) = 0.0001048576








The Cumulative Distribution Function, CDF,


F(X) = P(X ≤ x) is:





x


∑ P(X = t) =


t = 0





P( X ≤ 0 ) = 0.006046618


P( X ≤ 1 ) = 0.0463574


P( X ≤ 2 ) = 0.1672898


P( X ≤ 3 ) = 0.3822806


P( X ≤ 4 ) = 0.6331033


P( X ≤ 5 ) = 0.8337614


P( X ≤ 6 ) = 0.9452381


P( X ≤ 7 ) = 0.9877054


P( X ≤ 8 ) = 0.9983223


P( X ≤ 9 ) = 0.9998951


P( X ≤ 10 ) = 1








1 - F(X) is:





n


∑ P(X = t) =


t = x





P( X ≥ 0 ) = 1


P( X ≥ 1 ) = 0.9939534


P( X ≥ 2 ) = 0.9536426


P( X ≥ 3 ) = 0.8327102


P( X ≥ 4 ) = 0.6177194


P( X ≥ 5 ) = 0.3668967


P( X ≥ 6 ) = 0.1662386


P( X ≥ 7 ) = 0.05476188


P( X ≥ 8 ) = 0.01229455


P( X ≥ 9 ) = 0.001677722


P( X ≥ 10 ) = 0.0001048576
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