# 5. Taking a cruise is a costly discretionary expense. In a recent year, the top fivecruise lines in the world had this many passengers:4,133,000 2,369,000 1,295,000 928,000 679,000Round your answers to the nearest integer.a. The computations will be easier to work if you view this problem in termsof thousands of passengers. Represent each number in terms of thousandsof passengers.b. What is the mean number of passengers for these five cruise lines? (Givethe full number.)c. What is the range? (Give the full number.)d. What is the standard deviation? (Give the full number.)

The numbers in thousands are: 4133, 2369, 1295, 928, 679. The mean is 1881 thousand passengers, the range is 3454 thousand passengers, and the standard deviation is approximately 1218 thousand passengers.

### Explanation:

Part A: To represent each number in terms of thousands of passengers, we simplify them as follows: 4,133,000 is 4,133 thousands of passengers, 2,369,000 is 2,369 thousands, 1,295,000 is 1,295 thousands, 928,000 is 928 thousands, and 679,000 is 679 thousands.

Part B: To calculate the mean number of passengers, you would add up all the passenger numbers and then divide by the number of values (5 in this case). This adds up to 9,404,000 passengers or, in terms of thousands, 9,404. Divided by 5, this gives a mean of 1,880,800 passengers, or 1,881 thousands of passengers.

Part C: The range is calculated by subtracting the smallest number from the largest. In this case, that would be 4,133,000 - 679,000 = 3,454,000, which is also 3,454 thousands of passengers.

Part D: Calculating the standard deviation involves multiple steps. First, for each value, subtract the mean and square this result. Then, calculate the mean of these squared differences. Finally, take the square root of this mean. Doing so, the standard deviation is approximate 1,217,982 passengers, or 1,218 thousands of passengers.

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## Related Questions

Assume f(x) = ax^2+ bx + c Which one is true? f is a parabola that opens sideways to the right. f is a parabola that opens sideways to the left. f is a parabola that opens downward. f is a parabola that opens upward.​

Step-by-step explanation:

f is a parabola that opens upward.

An insurance company writes policies for a large number of newly-licensed drivers each year. Suppose 40% of these are low-risk drivers, 40% are moderate risk, and 20% are high risk. The company has no way to know which group any individual driver falls in when it writes the policies. None of the low-risk drivers will have an at-fault accident in the next year, but 10% of the moderate-risk and 20% of the high-risk drivers will have such an accident. If a driver has an at-fault accident in the next year, what is the probability that he or she is high-risk

The probability that he or she is high-risk is 0.50

Step-by-step explanation:

P(Low risk) = 40% = 0.40

P( Moderate risk) = 40% = 0.40

P(High risk) = 20% = 0.20

P(At - fault accident | Low risk) = 0% = 0

P(At-fault accident | Moderate risk) = 10% = 0.10

P(At-fault accident | High risk) = 20% = 0.20

If a driver has an at-fault accident in the next year, what is the probability that he or she is high-risk. Hence, We need to  calculate P( High risk | at-fault accident) = ?

Using Bayes' conditional probability theorem

P( High risk | at-fault accident) = ( P( High risk) * P(At-fault accident | High risk) ) /  { P( Low risk) * P(At-fault accident | Low risk) +P( Moderate risk) * P(At-fault accident | Moderate risk) +  P( High risk) * P(At-fault accident | High risk) }

P( High risk | at-fault accident)= (0.20 * 0.20) / ( 0.40 * 0 + 0.40 * 0.10 + 0.20 * 0.20 )

P( High risk | at-fault accident) = 0.04 / 0 + 0.04 + 0.04

P( High risk | at-fault accident) = 0.04 / 0.08

P( High risk | at-fault accident) = 0.50.

The probability that a driver is high-risk given that they had an at-fault accident can be found using Bayes' theorem. Given the probabilities provided in the question, the probability is approximately 0.3333 or 33.33%.

### Explanation:

To find the probability that a driver is high-risk given that they had an at-fault accident, we can use Bayes' theorem. Let's define the events:

1. A: Driver is high-risk
2. B: Driver has an at-fault accident

We are given the following probabilities:

1. P(A) = 0.20 (probability of a driver being high-risk)
2. P(B|A) = 0.20 (probability of an at-fault accident given that they are high-risk)
3. P(~A) = 0.80 (probability of a driver not being high-risk)
4. P(B|~A) = 0.10 (probability of an at-fault accident given that they are not high-risk)

Using Bayes' theorem, the probability of a driver being high-risk given that they had an at-fault accident is:

P(A|B) = (P(A) * P(B|A)) / (P(A) * P(B|A) + P(~A) * P(B|~A))

Substituting the given probabilities:

P(A|B) = (0.20 * 0.20) / (0.20 * 0.20 + 0.80 * 0.10) = 0.04 / (0.04 + 0.08) = 0.04 / 0.12 = 0.3333.

Therefore, the probability that a driver is high-risk given that they had an at-fault accident in the next year is approximately 0.3333 or 33.33%.

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If a elevator traveled 10 floors in 5 seconds , how many floors did it travel in 1 second ?

2

Step-by-step explanation:

it's division 10/5 is 2 which means in one second it travels 2 floors

5/5 = 1
10/5 = 2

2 floors/second

Nancy put 3 cups of vegetables in each container.

Step-by-step explanation:

1. Add the amount of vegetables together

2. Divided the number of vegetables by 5, since they are put into 5 separate containers.

____________________________________________________________2+1+2+1+3+6 = 15

15 divided by 5 or 5 x ? = 15

2 1/2 cups of vegetables

Step-by-step explanation:

This is an averages question, so add all numbers and divide by amount of numbers

2+1+2+1+3+6=15

15/6=2 1/2

Part A: Sam rented a boat at \$225 for 2 days. If he rents the same boat for 5 days, he has to pay a total rent of \$480. Write an equation in the standard form to represent the total rent (y) that Sam has to pay for renting the boat for x days.

Part B: Write the equation obtained in Part A using function notation.

Part C: Describe the steps to graph the equation obtained above on the coordinate axes. Mention the labels on the axes and the intervals.

First find the slope using the two points (2,225) and (5,480)
(480-225)/(5-2) = 85
Then find b.
225 = 85*2 + b
225 = 170 + b
55 = b
y = 85x + 55 OR f(x) = 85x + 55 This is the answer to PART B.
Standard form PART A is 85x - y = 55
PART C Plot the y intercept, 55. From the y intercept count the slope 85 up right 1 and plot a point, draw a line through the points
The x axis is the number of days so it would be in intervals of 1.
The y axis is money, I would do intervals of \$5

The white "Spirit" black bear (or Kermode) Ursus americanus kermodei, differs from the ordinary black bear by a single amino acid change in the melanocortin 1 receptor gene (MC1R). In this population, the gene has two forms (or alleles): the "white" allele b and the "black" allele B. The trait is recessive: white bears have two copies of the white allele of this gene (bb), whereas a bear is black if it has one or two copies of the black allele (Bb or BB). Both color morphs and all three genotypes are found together in the bear population of the northwest coast of British Columbia. If possessing the white allele has no effect on growth, survival, reproductive success, or mating patterns of individual bears, then the frequency of individuals with 0, 1, or 2 copies of the white allele (b) in the population will follow a binomial distribution. To investigate, Hedrick and Ritland (2011) sampled and genotype 87 bears from the northwest coast:42 were BB
24 were Bb
21 were bb
Assume that this is a random sample.
A formal hypothesis test was carried out to compare the observed and expected frequencies of genotypes. The procedure obtained P = 0.0001.
1. "The frequency distribution of genotypes has a binomial distribution in the population" is the________ hypothesis, whereas "The frequency distribution of genotypes does not have a binomial distribution" is the _________ hypothesis.
2. The degrees of freedom for the test statistic are __________.
Say whether the each of the following statements is true or false solely on the basis of these results:
3. The difference between the observed and expected frequencies is statistically significant.________
4. The test statistic exceeds the critical value corresponding to α = 0.05. ___________
5. The test statistic exceeds the critical value corresponding to α = 0.01. ____________

1)  i ) Null hypothesis, ii) Alternate  hypothesis

2)  Degree of Freedom = 14.9011

3) True

4)  True

5)  True

Step-by-step explanation:

1) "The frequency distribution of genotypes has a binomial distribution in the population" is the NULL hypothesis, whereas "The frequency distribution of genotypes does not have a binomial distribution" is the ALTERNATE hypothesis.

2)  Degrees of freedom: see attached calculation  for

3) True, the difference between the observed and expected frequencies is statistically significant since p-value < alpha(0.05)

4)  True

when, 14.9011 > 3.8415    where ∝ = 0.05

when, p-value < alpha,TS > critical value

here p-value < 0.0001 < 0.05

5)  True

when, 14.9011 > 6.635    where ∝ = 0.01

when, p-value < alpha,TS > critical value

here p-value < 0.0001 < 0.01

The null hypothesis states that the genotype frequency distribution follows a binomial distribution, with the alternative hypothesis stating the contrary. The degrees of freedom for the test statistic are two. The P value of 0.0001 is statistically significant, thus the test statistic exceeds the critical values corresponding to both α = 0.05 and α = 0.01.

### Explanation:

1. The frequency distribution of genotypes has a binomial distribution in the population is the null hypothesis, whereas The frequency distribution of genotypes does not have a binomial distribution is the alternative hypothesis.

2. For this chi-square test, since there are 3 possible outcomes (BB, Bb, bb) the degrees of freedom are 3-1 = 2.

3. The difference between the observed and expected frequencies is statistically significant, that is true. A P value of 0.0001 is highly significant, clearly less than 0.05, and denotes a significant difference.

4. The test statistic exceeds the critical value corresponding to α = 0.05. This is true, as the statistically significant low P value indicates the test statistic is in the critical region.

5. The test statistic exceeds the critical value corresponding to α = 0.01. This is also true.