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sdas [7]
3 years ago
10

The mean number of visitors at a national park in one weekend is 52. Assume the variable follows a poisson distribution. Find th

e probability that there will be 58 visitors at this park in one weekend. That is, find P(X=58)
Mathematics
1 answer:
dem82 [27]3 years ago
5 0

Answer:

0.0374722

Step-by-step explanation:

Given that :

μ = 52

x = 58

P(x, μ) = (e^-μ) * (μ^x)/ x!

P(58, 52) = ((e^-52) * (52^58)) / 58!

P(58, 52) = [(2.6102E−23 * 3.37437E99) / 2.35056E78]

P(58, 52) = 8.80807E76 / 2.35056E78

P(58, 52) = 0.0374722

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(6-3)(2/5)=
(3)(0.4)=
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Explain how to find the surface area of a 3-D figure
german
Well, for any prism, it is the lateral area + the area of two ends.  
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3 years ago
Draw a picture of the standard normal curve and shade the area that corresponds to the requested probabilities. Then use the sta
elena-14-01-66 [18.8K]

Answer:

a)P [ z > 1,38 ] = 0,08379

b) P [ 1,233 < z < 2,43 ]  = 0,1012

c)  P [ z > -2,43 ]  = 0,99245

Step-by-step explanation:

a) P [ z > 1,38 ] = 1 -  P [ z < 1,38 ]

From z-table  P [ z < 1,38 ] = 0,91621

P [ z > 1,38 ] = 1 - 0,91621

P [ z > 1,38 ] = 0,08379

b)  P [ 1,233 - 2,43 ]  must be  P [ 1,233 < z < 2,43 ]

P [ 1,233 < z < 2,43 ]  = P [ z < 2,43 ] - P [ z > 1,233 ]

P [ z < 2,43 ]  = 0,99245

P [ z > 1,233 ] = 0,89125    ( approximated value  without interpolation)

Then

P [ 1,233 < z < 2,43 ]  = 0,99245 - 0,89125

P [ 1,233 < z < 2,43 ]  = 0,1012

c) P [ z > -2,43 ]

Fom z-table

P [ z > -2,43 ] = 1 - P [ z < -2,43 ]

P [ z > -2,43 ] = 1 - 0,00755

P [ z > -2,43 ]  = 0,99245

8 0
3 years ago
The temperature in the late afternoon
Fantom [35]

Answer:

-7.5 - (8.5 + 5)

Step-by-step explanation:

We just add together the values of how much it dropped at different parts of the day, and subtract them from what we started with. The parenthesis are important cause we want to know what the total temperature drop for the day was before we subtract it to get the actual temperature we end up with.

6 0
3 years ago
Read 2 more answers
A billing company that collects bills for​ doctors' offices in the area is concerned that the percentage of bills being paid by
omeli [17]

Answer:

1) A. H0: p = 0.30

HA: p not equal to 0.30

2) A. The Independence Assumption is met.

C. The Randomization Condition is met.

D. The Success/Failure Condition is met.

3) Test statistic z = 2.089

P-value = 0.0367

4) C. Reject H0. There is sufficient evidence to suggest that the percentage of bills paid by medical insurance has changed.

Step-by-step explanation:

1) This is a hypothesis test for a proportion.

The claim is that there is a significant change in the percent of bills being paid by medical​ insurance.

As we are looking for evidence of a difference, no matter if it is higher or lower than the null hypothesis proportion, the alternative hypothesis is defined by a unequal sign.

Then, the null and alternative hypothesis are:

H_0: \pi=0.3\\\\H_a:\pi\neq 0.3

2) Cheking the conditions:

The independence assumption and the randomization condition are met as the bills were selected randomly from the population.

The 10% condition can not be checked, as we do not know the size of the population.

The success/failure condition is met as the products np and n(1-p) are bigger than 10 (the number of successes and failures are both bigger than 10).

3) The significance level is assumed to be 0.05.

The sample has a size n=9260.

The sample proportion is p=0.31.

 

The standard error of the proportion is:

\sigma_p=\sqrt{\dfrac{\pi(1-\pi)}{n}}=\sqrt{\dfrac{0.3*0.7}{9260}}\\\\\\ \sigma_p=\sqrt{0.000023}=0.005

Then, we can calculate the z-statistic as:

z=\dfrac{p-\pi-0.5/n}{\sigma_p}=\dfrac{0.31-0.3-0.5/9260}{0.005}=\dfrac{0.01}{0.005}=2.089

This test is a two-tailed test, so the P-value for this test is calculated as:

\text{P-value}=2\cdot P(z>2.089)=0.0367

As the P-value (0.0184) is smaller than the significance level (0.05), the effect is  significant.

The null hypothesis is rejected.

There is enough evidence to support the claim that there is a significant change in the percent of bills being paid by medical​ insurance.

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