Respuesta :
Answer:
a) [tex]P(X=41)=(50C41)(0.65)^{41} (1-0.65)^{50-41}=0.00421[/tex]
b) [tex]P(X=36)=(50C36)(0.65)^{36} (1-0.65)^{50-36}=0.0714[/tex]
[tex]P(X=37)=(50C37)(0.65)^{37} (1-0.65)^{50-37}=0.0502[/tex]
[tex]P(X=38)=(50C38)(0.65)^{38} (1-0.65)^{50-38}=0.0319[/tex]
And adding these values we got:
[tex] P(36 \leq X \leq 38)= 0.1535[/tex]
c) We can find the expected value given by:
[tex] E(X) = np =50*0.65 = 32.5[/tex]
And the standard deviation would be:
[tex] \sigma = \sqrt{np(1-p)} \sqrt{50*0.65*(1-0.65)}= 3.373[/tex]
We can use the approximation to the normal distribution and we have at leat 95% of the data within 2 deviations from the mean. And the lower limit for this case would be:
[tex] \mu -2\sigma = 32.5- 2*3.373 = 25.75[/tex]
And then we can consider a value of 18 as unusual lower for this case.
Step-by-step explanation:
Let X the random variable of interest "number cleared by arrest or exceptional", on this case we can model this variable with this distribution:
[tex]X \sim Binom(n=50, p=0.65)[/tex]
The probability mass function for the Binomial distribution is given as:
[tex]P(X)=(nCx)(p)^x (1-p)^{n-x}[/tex]
Where (nCx) means combinatory and it's given by this formula:
[tex]nCx=\frac{n!}{(n-x)! x!}[/tex]
Part a
We want this probability:
[tex]P(X=41)=(50C41)(0.65)^{41} (1-0.65)^{50-41}=0.00421[/tex]
Part b
We want this probability:
[tex] P(36 \leq X \leq 38)[/tex]
We can find the individual probabilities:
[tex]P(X=36)=(50C36)(0.65)^{36} (1-0.65)^{50-36}=0.0714[/tex]
[tex]P(X=37)=(50C37)(0.65)^{37} (1-0.65)^{50-37}=0.0502[/tex]
[tex]P(X=38)=(50C38)(0.65)^{38} (1-0.65)^{50-38}=0.0319[/tex]
And adding these values we got:
[tex] P(36 \leq X \leq 38)= 0.1535[/tex]
Part c
We can find the expected value given by:
[tex] E(X) = np =50*0.65 = 32.5[/tex]
And the standard deviation would be:
[tex] \sigma = \sqrt{np(1-p)} \sqrt{50*0.65*(1-0.65)}= 3.373[/tex]
We can use the approximation to the normal distribution and we have at leat 95% of the data within 2 deviations from the mean. And the lower limit for this case would be:
[tex] \mu -2\sigma = 32.5- 2*3.373 = 25.75[/tex]
And then we can consider a value of 18 as unusual lower for this case.
