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Answer:
The probability content for the confidence interval (3.49, 3.69) is 96%.
Step-by-step explanation:
The sample selected to determine the mean number of times the pet owners visited their veterinarian each year is of size, n = 475.
The sample selected is quite large, i.e. n > 30.
According to the Central Limit Theorem if we have a population with mean μ and standard deviation σ and we take appropriately huge random samples (n ≥ 30) from the population with replacement, then the distribution of the sample mean will be approximately normally distributed.
Then, the mean of the distribution of sample mean is given by,
[tex]\mu_{\bar x}=\mu\approx\bar x[/tex] ; for n → ∞.
And the standard deviation of the distribution of sample means is given by,
[tex]\sigma_{\bar x}=\frac{\sigma}{\sqrt{n}}\approx \frac{s}{\sqrt{n}}[/tex] ; for n → ∞.
So, the random variable X, defined as the number of visits to the veterinarian each year, follows a Normal distribution.
The (1 - α)% confidence interval for the population mean (μ) is:
[tex]CI=\bar x\pm z_{\alpha/2}\times \frac{\sigma}{\sqrt{n}}[/tex]
The information provided is:
[tex]\bar x=3.59\\s=1.045\\[/tex]
The confidence interval is (3.49, 3.69).
The margin of error of the confidence interval for the population mean is:
[tex]MOE=\frac{UL-LL}{2}= z_{\alpha/2}\times \frac{\sigma}{\sqrt{n}}[/tex]
Compute the MOE as follows:
[tex]MOE=\frac{UL-LL}{2}=\frac{3.69-3.49}{2}=0.10[/tex]
Compute the critical value of z as follows:
[tex]MOE=z_{\alpha/2}\times \frac{s}{\sqrt{n}}\\0.10=z_{\alpha/2}\times \frac{1.045}{\sqrt{475}}\\z_{\alpha/2}=2.0856\\z_{\alpha/2}\approx2.09[/tex]
Compute the value of [tex]P(-z_{\alpha/2}<Z<z_{\alpha/2})[/tex] as follows:
[tex]P(-z_{\alpha/2}<Z<z_{\alpha/2})=P(-2.09<Z<2.09)[/tex]
[tex]=P(Z<2.09)-P(Z<-2.09)\\=P(Z<2.09)-[1-P(Z<2.09)]\\=2P(Z<2.09)-1\\=(2\times 0.9817)-1\\=0.9634\\\approx0.96[/tex]
Thus, the probability content for the confidence interval (3.49, 3.69) is 96%.
Since the sample size is big enough ( > 30), the distribution can be approximated by normal distribution. You can convert the distribution to the standard normal distribution and then can use the z tables to find the p values.
The probability content for the given interval is 0.9586
How to form standard normal distribution?
Let the random variable X tracks the number of times the respondents visit their veterinarians each year. Then according to the given data, we have:
[tex]X \sim N(mean, std. dev) = N(3.59, 1.045)[/tex]
The distribution [tex]Y \sim N(a,b)[/tex] can be converted to standard normal distribution(the distribution with mean 0 and standard deviation as 1) [tex]Z \sim N(0,1)[/tex] by
[tex]Z = \dfrac{Y - \overline{Y}}{s/\sqrt{n}}[/tex]
where s denotes the standard deviation of Y's distribution and n denotes sample size (for population variates, we don't divide standard deviation by square root of sample size)
Thus, the standard variate for given sample will be:
[tex]Z = \dfrac{X - 3.59}{1.045/\sqrt{457}} = \dfrac{X-3.59}{0.049}[/tex]
The needed probability can be converted to probability of standard variate and then can be calculated as
[tex]P(3.49 < X < 3.69) = P( \dfrac{3.49 - 3.59}{0.049} < Z < \dfrac{3.69-3.59}{0.049})\\\\= P(-2.04 < Z < 2.04) = P(Z = 2.04) - P(Z = -2.04)\\= 0.9793-0.0207 = 0.9586[/tex]
Thus,
The probability content for the given interval is 0.9586
Learn more about standard normal distribution here:
https://brainly.com/question/14989264
