In a large accounting firm, the proportion of accountants with MBA degrees and at least five years of professional experience is
1 answer:
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
= 1 - P(A)
= 1 - 0.35
= 0.65
= 0.45
P(B) = 1 -
P(B) = 1 - 0.45
P(B) = 0.55
P(A ∩ B ) = 0.75
P(A ∩ B ) = 0.75 [ 1 - P(A ∪ B) ] because =
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) =
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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Using the distributions, it is found that there is a:
a) 0.1762 = 17.62% probability that 10 of the tosses will fall heads and 10 will fall tails.
b) 0% probability that 10 of the tosses will fall heads and 10 will fall tails.
c) 0.1742 = 17.42% probability that 10 of the tosses will fall heads and 10 will fall tails.
Item a:
Binomial probability distribution
The parameters are:
- x is the number of successes.
- n is the number of trials.
- p is the probability of a success on a single trial.
In this problem:
- 20 tosses, hence
.
- Fair coin, hence
.
The probability is <u>P(X = 10)</u>, thus:
0.1762 = 17.62% probability that 10 of the tosses will fall heads and 10 will fall tails.
Item b:
In a normal distribution with mean and standard deviation
, the z-score of a measure X is given by:
- It 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, which is the percentile of X.
- The binomial distribution is the probability of <u>x successes on n trials, with p probability</u> of a success on each trial. It can be approximated to the normal distribution with
.
The probability of an exact value is 0, hence 0% probability that 10 of the tosses will fall heads and 10 will fall tails.
Item c:
For the approximation, the mean and the standard deviation are:
Using continuity correction, this probability is , which is the <u>p-value of Z when X = 10.5 subtracted by the p-value of Z when X = 9.5.</u>
X = 10.5:
has a p-value of 0.5871.
X = 9.5:
has a p-value of 0.4129.
0.5871 - 0.4129 = 0.1742.
0.1742 = 17.42% probability that 10 of the tosses will fall heads and 10 will fall tails.
A similar problem is given at brainly.com/question/24261244