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Marina86 [1]
2 years ago
12

The histogram shows the number of vacation days taken by employees in the past year. Based on the histogram, which statement is

true?
A.
Four employees took 20 vacation days in the past year.

B.
Four employees took 25 vacation days in the past year.

C.
Four employees took between 20 and 25 vacation days in the past year.

D.
Between 20 and 25 employees took four vacation days in the past year.
Mathematics
1 answer:
Serhud [2]2 years ago
8 0
The correct answer is b
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Rocky Mountain National Park is a popular park for outdoor recreation activities in Colorado. According to U.S. National Park Se
Ugo [173]

Answer:

a) 0.6628 = 66.28% probability that at least 85 visitors had a recorded entry through the Beaver Meadows park entrance

b) 0.5141 = 51.41% probability that at least 80 but less than 90 visitors had a recorded entry through the Beaver Meadows park entrance

c) 0.5596 = 55.96% probability that fewer than 12 visitors had a recorded entry through the Grand Lake park entrance.

d) 0.9978 = 99.78% probability that more than 55 visitors have no recorded point of entry

Step-by-step explanation:

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

E(X) = np

The standard deviation of the binomial distribution is:

\sqrt{V(X)} = \sqrt{np(1-p)}

Normal probability distribution

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score 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. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

When we are approximating a binomial distribution to a normal one, we have that \mu = E(X), \sigma = \sqrt{V(X)}.

In this problem, we have that:

n = 175

(a) What is the probability that at least 85 visitors had a recorded entry through the Beaver Meadows park entrance?

46.7% of visitors to Rocky Mountain National Park in 2018 entered through the Beaver Meadows. This means that p = 0.467. So

\mu = E(X) = np = 175*0.467 = 81.725

\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{175*0.467*0.533} = 6.6

This probability, using continuity correction, is P(X \geq 85 - 0.5) = P(X \geq 84.5), which is 1 subtracted by the pvalue of Z when X = 84.5. So

Z = \frac{X - \mu}{\sigma}

Z = \frac{84.5 - 81.725}{6.6}

Z = 0.42

Z = 0.42 has a pvalue of 0.6628.

66.28% probability that at least 85 visitors had a recorded entry through the Beaver Meadows park entrance.

(b) What is the probability that at least 80 but less than 90 visitors had a recorded entry through the Beaver Meadows park entrance?

Using continuity correction, this is P(80 - 0.5 \leq X <  90 - 0.5) = P(79.5 \leq X \leq 89.5), which is the pvalue of Z when X = 89.5 subtracted by the pvalue of Z when X = 79.5. So

X = 89.5

Z = \frac{X - \mu}{\sigma}

Z = \frac{89.5 - 81.725}{6.6}

Z = 1.18

Z = 1.18 has a pvalue of 0.8810.

X = 79.5

Z = \frac{X - \mu}{\sigma}

Z = \frac{79.5 - 81.725}{6.6}

Z = -0.34

Z = -0.34 has a pvalue of 0.3669.

0.8810 - 0.3669 = 0.5141

51.41% probability that at least 80 but less than 90 visitors had a recorded entry through the Beaver Meadows park entrance

(c) What is the probability that fewer than 12 visitors had a recorded entry through the Grand Lake park entrance?

6.3% of visitors entered through the Grand Lake park entrance, which means that p = 0.063

\mu = E(X) = np = 175*0.063 = 11.025

\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{175*0.063*0.937} = 3.2141

This probability, using continuity correction, is P(X < 12 - 0.5) = P(X < 11.5), which is the pvalue of Z when X = 11.5. So

Z = \frac{X - \mu}{\sigma}

Z = \frac{11.5 - 11.025}{3.2141}

Z = 0.15

Z = 0.15 has a pvalue of 0.5596.

55.96% probability that fewer than 12 visitors had a recorded entry through the Grand Lake park entrance.

(d) What is the probability that more than 55 visitors have no recorded point of entry?

22.7% of visitors had no recorded point of entry to the park. This means that p = 0.227

\mu = E(X) = np = 175*0.227 = 39.725

\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{175*0.227*0.773} = 5.54

Using continuity correction, this probability is P(X \leq 55 + 0.5) = P(X \leq 55.5), which is the pvalue of Z when X = 55.5. So

Z = \frac{X - \mu}{\sigma}

Z = \frac{55.5 - 39.725}{5.54}

Z = 2.85

Z = 2.85 has a pvalue of 0.9978

0.9978 = 99.78% probability that more than 55 visitors have no recorded point of entry

8 0
2 years ago
HURRY BRAINLIEST AND 40 POINTS!!!
grandymaker [24]

Answer:

The answer is the first one and the last one.

Step-by-step explanation:

If an account cost $1 and each song is 1.5x. It has to be less than or equal to 25. So it would be $1 for (the account plus) 1.5x for (how many songs he buys) is less than or equal to 25 (the ammount on the gift card).

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Circle C has radius of 10 cm. Each of points B and D is on the midpoint of the radius. Find the area of the shaded region.
olya-2409 [2.1K]
Simple :)
Area of shaded part = are of 1/4 circle - area of both triangles.
Are of circle = pie r^2 so 100x3.14 = 314 cm2.
Area of triangle AOB= area of triangleDOE = bh/2= 5x10/2= 25 each
However, the traingles share a common area which is quad DOB(I)
Lets take traingle AOE, whose are is bh/2=10x10/2=50cm2.
50-area of triangle A(I)E= 50-(
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Which is the pattern for factoring the sum of cubes​
vovikov84 [41]

Answer: Its B

Step-by-step explanation: trust me i had this test

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