For a 95% confidence interval, the corresponding z-score is 1.96. Therefore the deviation will by 1.96*0.5 lbs = 0.98 lbs. Therefore, the confidence interval will be (5 - 0.98, 5 + 0.98), which is (4.02, 5.98). The weight range is from 4.02 lbs to 5.98 lbs.

5 0

3 years ago

What is x? 6(x+4)+3x-2

valina [46]

That is impossible to answer because this is an expression. you can only solve for x in equations

6 0

2 years ago

What is the square root of 36?

Ket [755]

6 is the answer noob

5 0

3 years ago

Read 2 more answers

The standard deviation for actuary salaries has a standard deviation of $36,730 . You collect a simple random sample of n=36 sal

Yanka [14]

Using the z-distribution, it is found that the lower limit of the 95% confidence interval is of $99,002.

<h3>What is a z-distribution confidence interval?</h3>

The confidence interval is:

$\overline{x} \pm z\frac{\sigma}{\sqrt{n}}$

In which:

$\overline{x}$ is the sample mean.

z is the critical value.

n is the sample size.

$\sigma$ is the standard deviation for the population.

In this problem, we have a 95% confidence level, hence $\alpha = 0.95$ , z is the value of Z that has a p-value of $\frac{1+0.95}{2} = 0.975$ , so the critical value is z = 1.96.

The other parameters are given as follows:

$\overline{x} = 111000, \sigma = 36730, n = 36$

Hence, the lower bound of the interval is:

$\overline{x} - z\frac{\sigma}{\sqrt{n}} = 111000 - 1.96\frac{36730}{\sqrt{36}} = 99002$

The lower limit of the 95% confidence interval is of $99,002.

More can be learned about the z-distribution at brainly.com/question/25890103

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8 0

2 years ago