Respuesta :
Answer:
71.08% probability that pˆ takes a value between 0.17 and 0.23.
Step-by-step explanation:
We use the binomial approxiation to the normal 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]p = 0.2, n = 200[/tex]. So
[tex]\mu = E(X) = np = 200*0.2 = 40[/tex]
[tex]\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{200*0.2*0.8} = 5.66[/tex]
In other words, find probability that pˆ takes a value between 0.17 and 0.23.
This probability is the pvalue of Z when X = 200*0.23 = 46 subtracted by the pvalue of Z when X = 200*0.17 = 34. So
X = 46
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{46 - 40}{5.66}[/tex]
[tex]Z = 1.06[/tex]
[tex]Z = 1.06[/tex] has a pvalue of 0.8554
X = 34
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{34 - 40}{5.66}[/tex]
[tex]Z = -1.06[/tex]
[tex]Z = -1.06[/tex] has a pvalue of 0.1446
0.8554 - 0.1446 = 0.7108
71.08% probability that pˆ takes a value between 0.17 and 0.23.
Probability helps us to know the chances of an event occurring. The probability that p takes a value between 0.17 and 0.23 is 71.08%.
What is Probability?
The probability helps us to know the chances of an event occurring.
[tex]\rm{Probability=\dfrac{Desired\ Outcomes}{Total\ Number\ of\ outcomes\ possible}[/tex]
As it is given that 20% of the adult women in the United States dye or highlight their hair, and the SRS is of size200(n). therefore, according to the Binomial distribution, the expected value, and the standard deviation of the sample can be written as,
[tex]\text{The expected value of the Sample}(\mu) = np = 200 \times 0.20 = 40[/tex]
[tex]\begin{aligned}\text{The standard deviation of the Sample}(\sigma) &= \sqrt{np(1-p)}\\ &= \sqrt{200 \times 0.20(1-0.20)}\\ &= \sqrt{200 \times 0.20 \times 0.80}\\&= \sqrt{32}\\&= 5.657\end{aligned}[/tex]
Now, using the standard normal distribution, the probability that p takes a value between 0.17 and 0.23 is the probability that the value of the X is between X₁ = 200 x 0.23 = 46 and X₂ = 200 x 0.17 = 34. Therefore, the probability that the value of the X is between 46 and 34 can be written as,
[tex]=P(X_2\leq X \leq X_1)\\\\= P(Z\leq \dfrac{X_1 - \mu}{\sigma}) - P(Z\leq \dfrac{X_2 - \mu}{\sigma})\\\\= P(Z\leq 1.06) - P(Z\leq -1.06)\\\\=0.8554 - 0.1446\\\\=0.7108\\\\=71.08\%[/tex]
Hence, the probability that p takes a value between 0.17 and 0.23 is 71.08%.
Learn more about Probability:
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