Mathematics College

A human resources manager keeps a record of how many years each employee at a large company has been working in their current role. The distribution of these years of experience are strongly skewed to the right with a mean of 333years and a standard deviation of 222 years. Suppose we were to take a random sample of 444employees and calculate the sample mean for their years of experience. We can assume independence between members in the sample.
What is the probability that the mean years of experience from the sample of 444 employees xˉxˉx, with, \bar, on top is greater than 3.53.53, point, 5 years?

Answers

Answer 1

Answer:

0.3085 = 30.85% probability that the mean years of experience from the sample of 4 is greater than 3.5 years.

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

When the distribution is normal, we use the z-score formula.

In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

The Z-score 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. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex].

For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean [tex]\mu = p[/tex] and standard deviation [tex]s = \sqrt{\frac{p(1-p)}{n}}[/tex]

Distribution of years of experience:

Mean 3, so [tex]\mu = 3[/tex]

Standard deviation 2, so [tex]\sigma = 2[/tex]

Sample of 4:

[tex]n = 4, s = \frac{2}{\sqrt{4}} = 1[/tex]

What is the probability that the mean years of experience from the sample of 4 is greater than 3.5?

1 subtracted by the pvalue of Z when X = 3.5. So

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

By the Central Limit Theorem

[tex]Z = \frac{X - \mu}{s}[/tex]

[tex]Z = \frac{3.5 - 3}{1}[/tex]

[tex]Z = 0.5[/tex]

[tex]Z = 0.5[/tex] has a pvalue of 0.6915

1 - 0.6915 = 0.3085

0.3085 = 30.85% probability that the mean years of experience from the sample of 4 is greater than 3.5 years.

Answer 2

Answer:

We cannot calculate this probability because the sampling distribution n ot normal

Step-by-step explanation: Khan academy

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which statement best describes the roots of f(x)=x^2-6x+25?

Answers

Answer:

permiteme veré en álgebra baldor

The roots of the equation f(x) = x² - 6x + 25 will be 3 - 4i and 3 + 4i.

What are the roots of the equation?

Let the equation be ax² + bx + c = 0. Then the roots of the equation will be

[tex]\rm x = \dfrac{-b \pm \sqrt{b^2 - 4 a c }}{2a}[/tex]

The quadratic equation is given below.

f(x) = x² - 6x + 25

Compare both the equations, then the value of a, b, and c will be 1, -6, and 25, respectively.

Then the roots of the equation are given as,

[tex]\rm x = \dfrac{-(-6) \pm \sqrt{(-6)^2 - 4 \times 1 \times 25 }}{2 \times 1}\\x = \dfrac{6 \pm \sqrt{-64}}{2}[/tex]

Simplify the equation further, then we have

x = 3 ± 4i

The roots of the equation f(x) = x² - 6x + 25 will be 3 - 4i and 3 + 4i.

More about the roots of the equation link is given below.

https://brainly.com/question/12029673

#SPJ2

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