Akbari.farahnaz

2022-01-04

1. Given a random sample size of n = 900 from a bi- nomial probability distribution with P = 0.10 do the following:

1. Find the probability that the number of successes is greater than 110.

2. Find the probability that the number of successes is fewer than 53.

3. Find the probability that the number of successes is between 55 and 120.

4. With probability 0.10, the number of successes is fewer than how many?

5. Withprobability0.08,thenumberofsuccessesis greater than how many?

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Since we only answer up to 3 sub-parts, we’ll answer the first 3.
Please resubmit the question and specify the other subparts (up to 3) you’d like answered.

Let us assume X=the random variable

Given that X follows a binomial distribution with n=900 and p=0.10

a) The probability that the number of successes is greater than 110 is:

$P\left(x>110\right)$

$=1-P\left(X\le 110\right)$

$=1-\text{binomcdf}\left(900,0.10,110\right)$

$=1-0.98685285379$

$=0.01314714621$

Answer(a): $0.01314714621$

b) The probability that the number of successes is fewer than 53 is:

$P\left(X<53\right)$

$=P\left(X\le 52\right)$

$=\text{binomcdf}\left(900,0.10,52\right)$

$=0.00000393759$

Answer(b):  $0.00000393759$

c) The probability that the number of successes is between 55 and 120 is:

$P\left(55\le X\le 120\right)$

$=P\left(X\le 120\right)-P\left(X<55\right)$

$=P\left(X\le 120\right)-P\left(X\le 54\right)$

$=\text{binomcdf}\left(900,0.10,120\right)-\text{binomcdf}\left(900,0.10,54\right)=0.99943052461-0.00001279394$

$=0.99941773067$

Answer(c):  $0.99941773067$

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