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3 years ago

The manager of a vacation resort believes the ages of adult visitors to the resort can be modeled by a normal distribution

Advanced Placement (AP)

1 answer:

hjlf3 years ago

8 0

The manager of a vacation resort believes that the ages of adult visitors to the resort can be modeled by a normal distribution. The manager surveyed a random sample of 765 visitors and recorded their age. A summary of the responses is shown in the frequency table, where x represents the age of the visitor.

(a) Construct a histogram of the distribution of ages.

(b) Write a few sentences to describe the distribution of ages of the adult visitors to the resort.

(c) Does the histogram provide convincing evidence that the surveyed ages come from a normal distribution? Explain your answer.

Age (years) Frequency

20≤x<30 160

30≤x<40. 130

40≤x<50 100

50≤x<60. 115

60≤x<70 125

70≤x<80 135

Answer:

Explanation:

a) Histogram is a basically a bar graph of the information. In this case the Histogram will have the ages on the bottom (horizontal x axis) and the heights of each bar equal to the number in that age bracket (frequency)... See attachment for histogram

b) once i have the histogram, I describe the general shape.

is it a bell shape (with the largest bar(s) in the middle and gets lower as you move away)?

Is it rectangular?

Is it like a side or a leaning bell curve?

It's shape gives the nature of the data which can be a good clue as to what the population looks like.

Since it is a inform of a stretched u-shape it is refers to nonlinear relationship between two variables, in particular, a dependent on an independent variable that is age depends on it frequency.

C. No because since the histogram curve is not a bell shape curve, then it cannot be normal distribution.

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Under the given assumptions, summer lasts 93 days (10 days for Jun, 31 days each for Jul and Aug, and 21 days for Sept), and winter lasts 90 days (11 days for Dec, 31 days for Jan, 28 days for Feb, and 20 days for Mar).

1. In the summer, we assume the Earth moves in a circular path with radius 94.5 million mi. In 365 days, the planet would cover a distance equal to the circumference of this circle, 2π * (94.5 million mi), or approximately (using π = 3.14) 593.46 million mi. Summer last 93 days, which is roughly 93/365 = 25.48% of one year. So during the summer, the Earth will traverse about 25.48% of the circumference of its circular path, or

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In the winter, the Earth's path is taken to be a circle of radius 91.4 million mi, so its circumference is 2π * (91.4 million mi) = 573.99 million mi. Winter has a duration of 90/365 = 24.68% of a year, so during winter the Earth travels

0.2468 * (91.4 million mi) = 22.6 million mi

2. The average speed of a body over some time interval is the distance the body travels divided by the duration of time.

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During the winter, it moves at a speed of

(22.6 million mi) / (90 days) = 0.251 million mi / day

3. Use the distances/arc lengths found in (1) to determine the measure of the central angle θ subtended by the summer and winter arcs. In either case, the ratio of arc length to circumference is equal to the ratio between the central angle and one complete revolution (2π radians).

By the same token, the ratio of sector area A to the area of the whole circle (πr^2) is proportional to the ratio of the central angle to 2π rad.

Summer:

(151.2 million mi) / (2π * (94.5 million mi)) = θ / (2π rad)

==> θ = 1.6 rad

A / (π * (94.5 million mi)^2) = (1.6 rad) / (2π rad)

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(22.6 million mi) / (2π * (91.4 million mi)) = θ / (2π rad)

==> θ = 0.2473 rad

A / (π * (91.4 million mi)^2) = (0.2473 rad) / (2π rad)

==> A = 1033.0 million square mi

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