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Harrizon [31]
3 years ago
15

Find the probability and interpret the results. If​ convenient, use technology to find the probability. The population mean annu

al salary for environmental compliance specialists is about ​$61 comma 500. A random sample of 30 specialists is drawn from this population. What is the probability that the mean salary of the sample is less than ​$57 comma 500​? Assume sigmaequals​$5 comma 800. The probability that the mean salary of the sample is less than ​$57 comma 500 is nothing.
Mathematics
1 answer:
Alisiya [41]3 years ago
4 0

Answer:0.004 is the probability that the mean salary of the sample is less than ​$57,500.

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = $60,500

Standard Deviation, σ = $6,400

Sample size, n  = 32

#if you need any queshtions answered within secs/mins hit me up and I got you!

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What is the slope of the line with the equation of y = 3/4 x that passes through the point (2, 1)?
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3 years ago
Suppose heights of seasonal pine saplings are normally distributed and have a known population standard deviation of 17 millimet
Mumz [18]

Answer:

296.693\leq x\leq 319.307

Step-by-step explanation:

The confidence interval for the population mean x can be calculated as:

x'-z_{\alpha /2}\frac{s}{\sqrt{n} } \leq x\leq x'+z_{\alpha /2}\frac{s}{\sqrt{n} }

Where x' is the sample mean, s is the population standard deviation, n is the sample size and z_{\alpha /2} is the z-score that let a proportion of \alpha /2 on the right tail.

\alpha is calculated as: 100%-99%=1%

So, z_{\alpha/2}=z_{0.005}=2.576

Finally, replacing the values of x' by 308, s by 17, n by 15 and z_{\alpha /2} by 2.576, we get that the confidence interval is:

308-2.576\frac{17}{\sqrt{15} } \leq x\leq 308+2.576\frac{17}{\sqrt{15} }\\308-11.307 \leq x\leq 308+11.307\\296.693\leq x\leq 319.307

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