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SIZIF [17.4K]
3 years ago
10

22% of all resumes received by a corporation for a management position are from females. 18 resumes will be received tomorrow.

Mathematics
1 answer:
Vladimir79 [104]3 years ago
3 0

Answer:

5.80% probability that exactly 1 resume will be from females.

Step-by-step explanation:

For each resume received by the corporation, there are only two possible outcomes. Either they are from a female, or they are not. The probability of a resume received being from a female is independent from other resumes. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

In which C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

And p is the probability of X happening.

22% of all resumes received by a corporation for a management position are from females.

This means that p = 0.22

18 resumes will be received tomorrow.

This means that n = 18

What is the probability that exactly 1 resume will be from females?

This is P(X = 1).

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 1) = C_{18,1}.(0.22)^{1}.(0.78)^{17} = 0.0580

5.80% probability that exactly 1 resume will be from females.

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Step-by-step explanation:

I find it convenient to start with a version of the point-slope form of the equation for a line. That is, for point (h, k) and slope m, ...

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For your m=3 and (h, k) = (3, 1), this equation becomes ...

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Answer:

Mean = (2.2 + 2.4 + 2.5 + 2.5 + 2.6 + 2.7)/6 = 2.48

Standard deviation = √(summation(x - mean)²/n

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Summation(x - mean)² = (2.2 - 2.48)^2 + (2.4 - 2.48)^2 + (2.5 - 2.48)^2 + (2.5 - 2.48)^2 + (2.6 - 2.48)^2 + (2.7 - 2.48)^2 = 0.1484

Standard deviation = √(0.1484/6

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Standard error = s/√n = 0.16/√6 = 0.065

Part B

Confidence interval is written as sample mean ± margin of error

Margin of error = z × s/√n

Since sample size is small and population standard deviation is unknown, z for 98% confidence level would be the t score from the student t distribution table. Degree of freedom = n - 1 = 6 - 1 = 5

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Confidence interval is 2.48 ± 0.22

Part C

We would set up the hypothesis test. This is a test of a single population mean since we are dealing with mean

For the null hypothesis,

H0: µ = 2.3

For the alternative hypothesis,

H1: µ > 2.3

This is a right tailed test

Since the number of samples is small and no population standard deviation is given, the distribution is a student's t.

Since n = 6

Degrees of freedom, df = n - 1 = 6 - 1 = 5

t = (x - µ)/(s/√n)

Where

x = sample mean = 2.48

µ = population mean = 2.3

s = samples standard deviation = 0.16

t = (2.48 - 2.3)/(0.16/√6) = 2.76

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Step-by-step explanation:

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