Respuesta :
Answer:
27.76% probability that a randomly selected bag of this size has 10 or more green candies
Step-by-step explanation:
I am going to use the normal approximation to the binomial to solve this question.
Binomial probability distribution
Probability of exactly x sucesses on n repeated trials, with p probability.
Can be approximated to a normal distribution, using the expected value and the standard deviation.
The expected value of the binomial distribution is:
[tex]E(X) = np[/tex]
The standard deviation of the binomial distribution is:
[tex]\sqrt{V(X)} = \sqrt{np(1-p)}[/tex]
Normal probability distribution
Problems of normally distributed samples can be solved using the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
When we are approximating a binomial distribution to a normal one, we have that [tex]\mu = E(X)[/tex], [tex]\sigma = \sqrt{V(X)}[/tex].
In this problem, we have that:
[tex]n = 40, p = 0.2[/tex]
So
[tex]\mu = E(X) = np = 40*0.2 = 8[/tex]
[tex]\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{40*0.2*0.8} = 2.53[/tex]
What is the probability that a randomly selected bag of this size has 10 or more green candies
Using continuity correction, this is [tex]P(X \geq 10 - 0.5) = P(X \geq 9.5)[/tex], which is 1 subtracted by the pvalue of Z when X = 9.5. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{9.5 - 8}{2.53}[/tex]
[tex]Z = 0.59[/tex]
[tex]Z = 0.59[/tex] has a pvalue of 0.7224
1 - 0.7224 = 0.2776
27.76% probability that a randomly selected bag of this size has 10 or more green candies
Answer:
[tex]P(x\geq 10)=0.2682[/tex]
Step-by-step explanation:
The number x of green candies in a bag of 40 candies follows a binomial distribution, because we have:
- n identical and independent events: 40 candies
- a probability p of success and (1-p) of fail: a probability of 0.2 to get a green candie and 0.8 to doesn't get a green candie.
So, the probability that in a bag of 40 candies, x are green is calculated as:
[tex]P(x)=\frac{n!}{x!(n-x)!}*p^{x}*(1-p)^{n-x}[/tex]
Replacing, n by 40 and p by 0.2, we get:
[tex]P(x)=\frac{40!}{x!(40-x)!}*0.2^{x}*(1-0.2)^{40-x}[/tex]
So, the probability that a randomly selected bag of this size has 10 or more green candies is equal to:
[tex]P(x\geq 10)=P(10)+P(11)+...+P(40)\\P(x\geq 10)=1-P(x<10)[/tex]
Where [tex]P(x<10)=P(0)+P(1)+P(2)+P(3)+P(4)+P(5)+P(6)+P(7)+P(8)+P(9)[/tex]
So, we can calculated P(0) and P(1) as:
[tex]P(0)=\frac{40!}{0!(40-0)!}*0.2^{0}*(1-0.2)^{40-0}=0.00013\\P(1)=\frac{40!}{1!(40-1)!}*0.2^{1}*(1-0.2)^{40-1}=0.00133[/tex]
At the same way, we can calculated P(2), P(3), P(4), P(5), P(6), P(7), P(8) and P(9) and get that P(x<10) is equal to:
[tex]P(x<10)=0.7318[/tex]
Finally, the probability [tex]P(x\geq 10)[/tex] that a randomly selected bag of this size has 10 or more green candies is:
[tex]P(x\geq 10)=1-P(x<10)\\P(x\geq 10)=1-0.7318\\P(x\geq 10)=0.2682[/tex]
