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Sagot :
Using the normal distribution and the central limit theorem, it is found that there is a 0.2033 = 20.33% probability that Becky’s proportion of poses held for over a minute is greater than Carla’s.
Normal Probability Distribution
In a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
- It 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, which is the percentile of X.
- By the Central Limit Theorem, for a proportion p in a sample of size n, the sampling distribution of sample proportion is approximately normal with mean [tex]\mu = p[/tex] and standard deviation [tex]s = \sqrt{\frac{p(1 - p)}{n}}[/tex], as long as [tex]np \geq 10[/tex] and [tex]n(1 - p) \geq 10[/tex].
For each sample, we have that:
[tex]p_B = 0.29, s_B = \sqrt{\frac{0.29(0.71)}{50}} = 0.046[/tex]
[tex]p_C = 0.35, s_C = \sqrt{\frac{0.35(0.65)}{50}} = 0.057[/tex]
For the distribution of the difference of the proportions of Becky and Carla, we have that:
[tex]\mu = p_B - p_C = 0.29 - 0.35 = 0.06[/tex]
[tex]s = \sqrt{s_B^2 + s_C^2} = \sqrt{0.046^2 + 0.057^2} = 0.073[/tex]
The probability that Becky’s proportion of poses held for over a minute is greater than Carla’s is the probability that the subtraction is above 0, that is, 1 subtracted by the p-value of Z when X = 0, hence:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
By the Central Limit Theorem
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{0 + 0.06}{0.073}[/tex]
[tex]Z = 0.82[/tex]
[tex]Z = 0.82[/tex] has a p-value of 0.7967.
1 - 0.7967 = 0.2033.
0.2033 = 20.33% probability that Becky’s proportion of poses held for over a minute is greater than Carla’s.
To learn more about the normal distribution and the central limit theorem, you can check https://brainly.com/question/24663213
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