# Please oblige me next to a statistics problem!?

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

A poll found that in the order of 40% of employees missed work due to a put money on injury. Let x be the number of sampled workers who enjoy missed work due to back injury.

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

b) Use the poll facts to estimate p for the binomial random changeable of part a.

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

d) From the example in element 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 posterior injury.

Answers:    a) The binomial is the sum of n independent and identically distributed Bernoulli trails. respectively work in the taste is a Bernoulli trial, either the did or did not miss work due to a support injury.

b) p = 0.40

c and c)

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

In general, if X have the binomial distribution with n trials and a nouns 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 plus 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 nouns and n - x failures.
Or, surrounded by 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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