A marketing research assistant for a veterinary hospital surveyed a random sample of 457 pet owners. Respondents were asked to indicate the number of times that they visit their veterinarian each year. e sample mean response was 3.59 and the sample standard deviation was 1.045. Based on these results, a condence interval from 3.49 to 3.69 was calculated for the population mean. Find the probability content for this interval

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,

; for n → ∞.

And the standard deviation of the distribution of sample means is given by,

; 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:

The information provided is:

The confidence interval is (3.49, 3.69).

The margin of error of the confidence interval for the population mean is:

Compute the MOE as follows:

Compute the critical value of z as follows:

Compute the value of as follows:

Thus, the probability content for the confidence interval (3.49, 3.69) is 96%.

hope it you

step-by-step explanation:

sin(7pi/6 - pi/3)  

= sin(7pi/6 - 2pi/6)  

= sin(5pi/6)  

= sin(pi - pi/6)  

= sin(pi/6)  

= 1/2  

sin 7pi/6 - sin pi/3  

= sin (pi + pi/6) - sin pi/3  

= -sin pi/6 - sin pi/3  

= -1/2 - sqrt(3)/2  

= -(1+sqrt(3))2

answer: 414,377,008

step-by-step explanation:

just multiply it