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Sagot :
Using the normal distribution and the central limit theorem, we have that:
a) The sampling distribution is approximately normal, with mean 0.22 and standard error 0.0338.
b) There is a 0.1867 = 18.67% probability that in a random sample of 150 couples more than 25% met online.
c) There is a 0.2584 = 25.84% probability that in a random sample of 150 couples between 15% and 20% met online.
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].
In this problem:
- 22% of couples meet online, hence p = 0.22.
- A sample of 150 couples is taken, hence n = 150.
Item a:
The mean and the standard error are given by:
[tex]\mu = p = 0.22[/tex]
[tex]s = \sqrt{\frac{p(1 - p)}{n}} = \sqrt{\frac{0.22(0.78)}{150}} = 0.0338[/tex]
The sampling distribution is approximately normal, with mean 0.22 and standard error 0.0338.
Item b:
The probability is one subtracted by the p-value of Z when X = 0.25, hence:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
By the Central Limit Theorem:
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{0.25 - 0.22}{0.0338}[/tex]
Z = 0.89
Z = 0.89 has a p-value of 0.8133.
1 - 0.8133 = 0.1867.
There is a 0.1867 = 18.67% probability that in a random sample of 150 couples more than 25% met online.
Item c:
The probability is the p-value of Z when X = 0.2 subtracted by the p-value of Z when X = 0.15, hence:
X = 0.2:
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{0.2 - 0.22}{0.0338}[/tex]
Z = -0.59
Z = -0.59 has a p-value of 0.2776.
X = 0.15:
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{0.15 - 0.22}{0.0338}[/tex]
Z = -2.07
Z = -2.07 has a p-value of 0.0192.
0.2776 - 0.0192 = 0.2584.
There is a 0.2584 = 25.84% probability that in a random sample of 150 couples between 15% and 20% met online.
To learn more about the normal distribution and the central limit theorem, you can check https://brainly.com/question/24663213
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