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Mandarinka [93]
3 years ago
8

In a large accounting firm, the proportion of accountants with MBA degrees and at least five years of professional experience is

75% as large as the proportion of accountants with no MBA degree and less than five years of professional experience. Furthermore, 35% of the accountants in this firm have MBA degrees, and 45% have fewer than five years of professional experience. If one of the firm's accountants is selected at random, what is the probability that this accountant has an MBA degree or at least five years of professional experience, but not both
Mathematics
1 answer:
Rasek [7]3 years ago
3 0

Answer:

The probability that this accountant has an MBA degree or at least five years of professional experience, but not both is 0.3

Step-by-step explanation:

From the given study,

Let A be the event that the accountant has an MBA degree

Let B be the event that the accountant has at least 5 years of professional experience.

P(A) = 0.35

P(A)^C = 1 - P(A)

P(A)^C = 1 - 0.35

P(A)^C = 0.65

P(B)^C = 0.45

P(B) = 1 - P(B)^C

P(B) = 1 - 0.45

P(B) = 0.55

P(A ∩ B ) = 0.75 P(A^C \ \cap  \  B^C)

P(A ∩ B ) = 0.75 [ 1 - P(A ∪ B) ]  because  P(A^C \ \cap  \  B^C)  = P(A \cup B)^C

SO;

P(A ∩ B ) = 0.75 [ 1 - P(A) - P(B) + P(A ∩ B) ]

P(A ∩ B ) = 0.75 [ 1 - 0.35 - 0.55 + P(A ∩ B) ]

P(A ∩ B ) - 0.75  P(A ∩ B) = 0.75 [1 - 0.35 -0.55 ]

0.25  P(A ∩ B) = 0.075

P(A ∩ B) = \dfrac{0.075}{0.25}

P(A ∩ B) = 0.3

The probability that this accountant has an MBA degree or at least five years of professional experience, but not both is: P(A ∪ B ) - P(A ∩ B)

= P(A) + P(B) - 2P( A ∩ B)

= (0.35 + 0.55) - 2(0.3)

= 0.9 - 0.6

= 0.3

∴

The probability that this accountant has an MBA degree or at least five years of professional experience, but not both is 0.3

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b) 0.8098

c) Between $24.75 and $27.25.

Step-by-step explanation:

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

Normal probability distribution

Problems of normally distributed samples are solved using the z-score formula.

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

Z = \frac{X - \mu}{\sigma}

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 \mu and standard deviation \sigma, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

In this problem, we have that:

\mu = 26, \sigma = 6, n = 62, s = \frac{6}{\sqrt{62}} = 0.762

(a)

What is the likelihood the sample mean is at least $27.00?

This is 1 subtracted by the pvalue of Z when X = 27. So

Z = \frac{X - \mu}{\sigma}

By the Central Limit Theorem

Z = \frac{X - \mu}{s}

Z = \frac{27 - 26}{0.762}

Z = 1.31

Z = 1.31 has a pvalue of 0.9049

1 - 0.9049 = 0.0951

(b)

What is the likelihood the sample mean is greater than $25.00 but less than $27.00?

This is the pvalue of Z when X = 27 subtracted by the pvalue of Z when X = 25. So

X = 27

Z = \frac{X - \mu}{s}

Z = \frac{27 - 26}{0.762}

Z = 1.31

Z = 1.31 has a pvalue of 0.9049

X = 25

Z = \frac{X - \mu}{s}

Z = \frac{25 - 26}{0.762}

Z = -1.31

Z = -1.31 has a pvalue of 0.0951

0.9049 - 0.0951 = 0.8098

c)Within what limits will 90 percent of the sample means occur?

50 - 90/2 = 5

50 + 90/2 = 95

Between the 5th and the 95th percentile.

5th percentile

X when Z has a pvalue of 0.05. So X when Z = -1.645

Z = \frac{X - \mu}{s}

-1.645 = \frac{X - 26}{0.762}

X - 26 = -1.645*0.762

X = 24.75

95th percentile

X when Z has a pvalue of 0.95. So X when Z = 1.645

Z = \frac{X - \mu}{s}

1.645 = \frac{X - 26}{0.762}

X - 26 = 1.645*0.762

X = 27.25

Between $24.75 and $27.25.

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