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statuscvo [17]
2 years ago
15

For each day that Sasha travels to work, the probability that she will experience a delay due to traffic is 0.2. Each day can be

considered independent of the other days. (b.) What is the probability that Sasha's first delay due to traffic will occur after the fifth day of travel to work? (c.) consider a random sample of 21 days that Sasha will travel to work. for the proportion of those days that she will experience a delay due to traffic, is the sampling distribution of the sampling proportion approximately normal? justify your answer.
Mathematics
1 answer:
Stolb23 [73]2 years ago
5 0

Answer:

a) P(X≥3) = 0.8213

b) The probability that Sasha's first delay due to traffic will occur after the fifth day of travel to work = 0.08192

c) This sampling distribution does not approximate a normal distribution.

Step-by-step explanation:

On a day that Sasha goes to work, let the probability that Sasha experience a delay due to traffic = P(T) = 0.2

The probability of no delay due to traffic = P(T') = 1 - 0.2 = 0.8

a) For the next 21 days that Sasha travels to work, what is the probability that Sasha will experience a delay due to traffic on at least 3 of the days?

This is a binomial distribution problem because a binomial experiment is one in which the probability of success doesn't change with every run or number of trials. It usually consists of a number of runs/trials with only two possible outcomes, a success or a failure.

The outcome of each trial/run of a binomial experiment is independent of one another.

Binomial distribution function is represented by

P(X = x) = ⁿCₓ pˣ qⁿ⁻ˣ

n = total number of sample spaces = 21

x = Number of successes required = ≥3

p = probability of success = 0.2

q = probability of failure = 0.8

P(X≥3) = 1 - P(X<3) = 1 - [P(X=0) + P(X=1) + P(X=2)] = 0.82129716679 = 0.8213

b) According to the question, each day can be considered independent of the other days

So, the probability that Sasha's first delay due to traffic will occur after the fifth day of travel to work mean that she doesn't experience delay due to traffic in the first 4 days.

Required probability = [P(T')]⁴ × P(T) = (0.8×0.8×0.8×0.8×0.2) = 0.08192

c) According to the Central limit theorem, a sample size large enough, selected from a population distribution that is random and independent, will approximate a normal distribution with Mean of sampling distribution (μₓ) equal to the population mean (μ) and the standard deviation of sampling distribution determined from the population standard deviation, the sample size and the population size.

But, how large is "large enough"?

A sample size of 30 is usually deemed large enough when the population distribution is roughly normal. But if the original population is distinctly not normal, the sample size should be be even larger.

And this sampling distribution does not satisfy any of the two statements above as 21 < 30.

Hence, this sampling distribution does not approximate a normal distribution.

Hope this Helps!!!

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Answer:

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3/150 = 1/50

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romanna [79]

Answer:

60°

Step-by-step explanation:

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8 0
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Ivahew [28]

One way to solve the system is to <u>substitute</u> a variable.

<u>Explanation:</u>

One approach to solve an equation is by substitution of one variable. Right now, a condition for one factor, at that point substitute that arrangement in the other condition, and explain. All value(s) of the variable(s) that fulfills a condition, disparity, arrangement of conditions, or arrangement of imbalances.

The technique for tackling "by substitution" works by settling one of the conditions (you pick which one) for one of the factors (you pick which one), and afterward stopping this go into the other condition, "subbing" for the picked variable and fathoming for the other. At that point you back-explain for the principal variable.

8 0
2 years ago
Read 2 more answers
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Paul [167]

Answer: A

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we have to divide it evenly in 9 pieces, so we divide $88.74 into 9 groups which then equals $9.86

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7 0
2 years ago
Merle Fonda opened a new savings account. She deposited $40,000 at 10% compounded semiannually. At the start of the fourth year,
IrinaVladis [17]
Use compound interest formula  F=P(1+i)^n twice, one for each deposit and sum the two results.

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For the P=20000, deposited at the START of the fourth year, which is the same as the end of the third year.
i=5% (semi-annual
n=2*(6-3), n = 6 
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F=P(1+i)^n = 20000(1+0.05)^6 = 26801.913

Total amount after 6 years
= 71834.253 + 26801.913
=98636.17   (to the nearest cent.)
8 0
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