Respuesta :
Answer:
a) 0.2741 = 27.41% probability that at least 13 believe global warming is occurring
b) 0.7611 = 76.11% probability that at least 110 believe global warming is occurring
Step-by-step explanation:
Binomial probability distribution
The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
In which [tex]C_{n,x}[/tex] is the number of different combinations of x objects from a set of n elements, given by the following formula.
[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]
And p is the probability of X happening.
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]p = 0.71[/tex]
(a) For a sample of 16 Americans, what is the probability that at least 13 believe global warming is occurring?
Here [tex]n = 16[/tex], we want [tex]P(X \geq 13)[/tex]. So
[tex]P(X \geq 13) = P(X = 13) + P(X = 14) + P(X = 15) + P(X = 16)[/tex]
In which
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 13) = C_{16,13}.(0.71)^{13}.(0.29)^{3} = 0.1591[/tex]
[tex]P(X = 14) = C_{16,14}.(0.71)^{14}.(0.29)^{2} = 0.0835[/tex]
[tex]P(X = 15) = C_{16,15}.(0.71)^{15}.(0.29)^{1} = 0.0273[/tex]
[tex]P(X = 16) = C_{16,16}.(0.71)^{16}.(0.29)^{0} = 0.0042[/tex]
[tex]P(X \geq 13) = P(X = 13) + P(X = 14) + P(X = 15) + P(X = 16) = 0.1591 + 0.0835 + 0.0273 + 0.0042 = 0.2741[/tex]
0.2741 = 27.41% probability that at least 13 believe global warming is occurring
(b) For a sample of 160 Americans, what is the probability that at least 110 believe global warming is occurring?
Now [tex]n = 160[/tex]. So
[tex]\mu = E(X) = np = 160*0.71 = 113.6[/tex]
[tex]\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{160*0.71*0.29} = 5.74[/tex]
Using continuity correction, this is [tex]P(X \geq 110 - 0.5) = P(X \geq 109.5)[/tex], which is 1 subtracted by the pvalue of Z when X = 109.5. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{109.5 - 113.6}{5.74}[/tex]
[tex]Z = -0.71[/tex]
[tex]Z = -0.71[/tex] has a pvalue of 0.2389
1 - 0.2389 = 0.7611
0.7611 = 76.11% probability that at least 110 believe global warming is occurring
