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. What is the probability that Sasha's first delay due to traffic will occur after the fifth day of travel to work?

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!!!

0.8213

Step-by-step explanation:

-This is a binomial probability problem given by the function:

Given that n=21 and p=0.2, the probability that she experience a delay on at least 3 days is calculated as:

Hence, the probability that she experience delay on at least 3 days is 0.8213

0.8213

Step-by-step explanation:

-This is a binomial probability problem given by the function:

Given that n=21 and p=0.2, the probability that she experience a delay on at least 3 days is calculated as:

Hence, the probability that she experience delay on at least 3 days is 0.8213

The answer is in the image 

For 1 flavor there are 9 topping

Therefore, for 5 different flavors there will be 5*9 choices

No of choices= 5*9

=45 

The answer is in the image 

The total nom of  code that can be used is equal to 5+3 = 8

y=2x+15

where y=Value of coin

x=Age in years

Value of coin after 19 years=2*19+15

=$53

Therefore, Value after 19 years=$53

F=ma

where F=force

m=mass

a=acceleration

Here,

F=4300

a=3.3m/s2

m=F/a

    =4300/3.3

    =1303.03kg

Approximately it is aqual to 1300kg