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dedylja [7]
3 years ago
7

During the dry month of August, one U.S. city has measurable rain on average only 4 days per month. Assume all months have 30 da

ys. *(Round your answers to 1 decimal place.) **(Round the intermediate values to 4 decimal places. Round your answer to 4 decimal places.) (a) If the arrival of rainy days is Poisson distributed in this city during the month of August, what is the average number of days that will pass between measurable rain? *
Mathematics
1 answer:
Doss [256]3 years ago
8 0

Answer:

The average number of days that will pass between measurable rain is 7.5.

Step-by-step explanation:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}

In which

x is the number of sucesses

e = 2.71828 is the Euler number

\mu is the mean in the given time interval.

The expected period between the event happening is \frac{1}{\mu}

During the dry month of August, one U.S. city has measurable rain on average only 4 days per month. Assume all months have 30 days.

This means that

\mu = \frac{4}{30}

(a) If the arrival of rainy days is Poisson distributed in this city during the month of August, what is the average number of days that will pass between measurable rain? *

\frac{1}{\frac{4}{30}} = \frac{30}{4} = 7.5

The average number of days that will pass between measurable rain is 7.5.

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Trapezoid ABCD is congruent to trapezoid A"B"C"D". Which sequence of transformations could have been used to transform trapezoid
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Answer:ygggydtdgdgdfddgfs

Step-by-step explanation:

8 0
3 years ago
This set of ordered pairs shows a relationship between x and y. {(0, -2), (3, 7), (6, 16), (6, 15), (8, 21), (10, 28), (11,31)}
Hunter-Best [27]

Answer:

The closest to the output when the input is approximately 12

Step-by-step explanation:

The given (x, y) coordinates are;

The line of best fit is

x, 0, 3, 6, 6, 8, 10, 11

y, -2, 7, 16, 15, 21, 28, 31

The line of best fit can be obtained from the scatter plot of the given data from where a linear pattern is apparent

A linear line of best fit is a trend line that gives an overall cumulative minimum distance of all the points from the line

The method for constructing a line of best fit includes;

1) The least Squares Method

2) The method of linear regression

3) Construction of line of best fit

a) The area method

b) The dividing method

The least squares equation is given as follows;

\hat y = a + b·x

b = \dfrac{\Sigma (x_i - \overline x) \cdot (y_i - \overline y)}{\Sigma (x_i - \overline x)^2 }

From MS Excel, we have;

{\Sigma (x_i - \overline x) \cdot (y_i - \overline y)}{ } = 266.8571

{\Sigma (x_i - \overline x)^2 } = 89.42857

∴ b = 266.8571/89.42857 ≈ 2.984

a =\overline y- b \cdot \overline x

From MS Excel, with the given data, we get;

\overline y = 16.57143

\overline x = 6.285714

Therefore;

a = 16.57143 - 2.984 × 6.285714 = -2.185

Therefore, we get the following regression equation;

\hat y = -2.185 + 2.9·x

Where;

x = The input

\hat y = The output

Therefore, when x = 5, we get;

\hat y = -2.185 + 2.9 × 5 = 12.315

Therefore, the closest to the output when the input is 5, y ≈ 12

8 0
3 years ago
. Suppose (as is roughly true) that 88% of college men and 82% of college women were employed last summer. A sample survey inter
denis23 [38]

Answer:

(a) The approximate distribution of the proportion pf of women who worked = 0.82, the standard distribution = 0.00738

The approximate distribution of the proportion pM of men who worked = 0.88, the standard distribution ≈ 0.01625

(b)  pM - pF =  0.06

While pF - pM = -0.06

The difference in the standard deviation ≈ 0.01025

(c) The probability that a higher proportion of women than men worked last year is 0

Step-by-step explanation:

(a) The given information are;

The percentage of college men that were employed last summer = 88%

The percentage of college women that were employed last summer = 82%

The number of college men interviewed in the survey = 400

The number of college women interviewed in the survey = 400

Therefore, given that the proportion of women that worked = 0.82, we have for the binomial distribution;

p = 0.82

q = 1 - 0.82 = 0.18

n = 400

Therefore;

p × n = 0.82 × 400 = 328 > 10

q × n = 0.18 × 400 = 72 > 10

Therefore, the binomial distribution is approximately normal

We have;

The \  mean = p = 0.82\\\\The \ standard \ deviation, \sigma  = \sqrt{ \dfrac{p \times q}{{n} } }= \sqrt{ \dfrac{0.82 \times 0.18}{{400} }} \approx  0.01921\\

Therefore, the approximate distribution of the proportion pf of women who worked = 0.82, the standard distribution ≈ 0.01921

Similarly, given that the proportion of male that worked = 0.88, we have for the binomial distribution;

p = 0.88

q = 1 - 0.88 = 0.12

n = 400

Therefore;

p × n = 0.88 × 400 = 352 > 10

q × n = 0.12 × 400 = 42 > 10

Therefore, the binomial distribution is approximately normal

We have;

The \  mean = p = 0.88\\\\The \ standard \ deviation, \sigma  = \sqrt{ \dfrac{p \times q}{{n} } }= \sqrt{ \dfrac{0.88 \times 0.12}{{400} }} \approx  0.01625\\

Therefore, the approximate distribution of the proportion pM of men who worked = 0.88, the standard distribution ≈ 0.01625

(b) Given two normal random variables, we have

The distribution of the difference the two normal random variable = A normal random variable

The mean of the difference = The difference of the two means = pM - pF = 0.88 - 0.82 = 0.06

While pF - pM = -0.06

The difference in the standard deviation, giving only the real values, is given as follows;

The \ difference \ in \ standard \ deviation  = \sqrt{ \dfrac{p_1 \times q_1}{{n_1} } -\dfrac{p_2 \times q_2}{{n_2} } }\\\\= \sqrt{\dfrac{0.82 \times 0.18}{{400} }-\dfrac{0.88 \times 0.12}{{400} }} \approx  0.01025\\

(c) When there is no difference between the the proportion of men and women that worked last summer, the probability that there is a difference = 0

Therefore, taking 0 as the standard score, we have;

z = \dfrac{0 - (-0.06)}{0.01025}  \approx 5.86

Given that the maximum values for a cumulative distribution table is approximately 4, we have that the probability that a higher proportion of women than men worked last year is 0.

3 0
3 years ago
Isa bought a number of pears and apples and paid $128 for them. If a pear costs $3.00 and an apple costs $4,00, how many apples
rodikova [14]

The numbers of pears and apples bought are 16 and 20 respectively

let

x = number of pears

y = number of apples

He bought 4 more apples than pears. Therefore,

  • y = x + 4

He bought a number of pears and apples and paid $128 for them. Therefore,

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<h3>System of equation:</h3>

y - x = 4

4y + 3x = 128

Therefore,

3y - 3x = 12

4y + 3x = 128

7y = 140

y = 140 / 7

y = 20

Therefore,

20 = x + 4

x = 20 - 4

x = 16

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Oksanka [162]

Answer:

-1.5

Step-by-step explanation:

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