Respuesta :
Answer:
1) [tex]P(X>64)=P(Z>-0.125)=1-P(Z<-0.125)=1-0.4503=0.5497[/tex]
2) [tex]P(X=2)=(4C2)(0.94)^2 (1-0.94)^{4-2}=0.0191[/tex]
Step-by-step explanation:
Previous concepts
Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".
The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".
The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".
Problem 1
Let X the random variable that represent the scores of a population, and for this case we know the distribution for X is given by:
[tex]X \sim N(65,8)[/tex]
Where [tex]\mu=65[/tex] and [tex]\sigma=8[/tex]
We are interested on this probability
[tex]P(X>64)[/tex]
And the best way to solve this problem is using the normal standard distribution and the z score given by:
[tex]z=\frac{x-\mu}{\sigma}[/tex]
[tex]P(X>64)=P(\frac{X-\mu}{\sigma}}>\frac{64-\mu}{\sigma}})=P(Z>-0.125)[/tex]
And we can find this probability on this way:
[tex]P(Z>-0.125)=1-P(Z<-0.125)=1-0.4503=0.5497[/tex]
Problem 2
Let X the random variable of interest, on this case we now that:
[tex]X \sim Binom(n=4, p=0.94)[/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]
And we want to find this probability:
[tex]P(X=2)=(4C2)(0.94)^2 (1-0.94)^{4-2}=0.0191[/tex]
