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Can Someone Help Me To Solve This Statistic Question It

Can someone help solve STATISTICS question?

Statistics
Mechanical failures of a certain type of aircraft constitute a Poisson process. On the average, based on past records, an aircraft will have a failure once every 10000 hours of flight time. If such aircrafts are scheduled for inspection and maintenance after every 5000 flight hours, what is the probability of mechanical failure of an aircraft between inspections?

Goodnight everyone , Can someone solve This Statistic Question ?

An internal study by the Technology Services department at Lahey Electronics revealed company employees receive an average of 4.0 non-work-related e-mails per hour. Assume the arrival of these e-mails is approximated by the Poisson distribution.


A - What is the probability Linda Lahey, company president, received exactly 2 non-work-related e-mails between 4 P.M. and 5 P.M. yesterday? (Round your probability to 4 decimal places.)


B- What is the probability she received 7 or more non-work-related e-mails during the same period? (Round your probability to 4 decimal places.)

C-What is the probability she did not receive any non-work-related e-mails during the period? (Round your probability to 4 decimal places.)

Can anyone help me solve this statistics question?

35,000*6=210000
210000- (37000+38000+33000+39000+29000)=$34,000

How can I get someone to come and help me with statistics?

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I have a few statistic problem that I am stuck on. Can someone help me out please?

15. Solve the problem.
A tourist in Ireland wants to visit six different cities. If the route is randomly selected, what Is the probability that the tourist will visit the cities in alphabetical order? Round your answer to three decimal places.

A. 0.016
B. 0.167
C. 0.001
D. 0.008

16. Solve the problem.
Nine guests are invited for dinner. How many ways can they be seated at a dinner table If the table is straight with seats only on one side?

A. 40,320
B. 5
C. 3,628,800
D. 362,880

17. Solve the problem.
If a couple has nine boys and two girls, how many gender sequences are possible?

A. 11
B. 8
C. 55
D. 16

Can someone tell me the steps I need to take to solve this statistics problem? Thanks?

Number of cases 10
Find the mean, add all of the observations and divide by 10
Mean 315.6
Find the standard deviation.
The values here are (an example)0,260,356,403,536,0,268,369,428,...
Squared deviations
(0-315.6)^2 = (-315.6)^2 = 99603.36
(260-315.6)^2 = (-55.6)^2 = 3091.36
(356-315.6)^2 = (40.4)^2 = 1632.16
(403-315.6)^2 = (87.4)^2 = 7638.76
(536-315.6)^2 = (220.4)^2 = 48576.16
(0-315.6)^2 = (-315.6)^2 = 99603.36
(268-315.6)^2 = (-47.6)^2 = 2265.76
(369-315.6)^2 = (53.4)^2 = 2851.56
(428-315.6)^2 = (112.4)^2 = 12633.76
(536-315.6)^2 = (220.4)^2 = 48576.16
Add the squaed deviations and divide by 9. That's the variance.
Variance = 326472.4/9
Variance 36274.7111
Standard deviation = sqrt(variance) =sqrt(326472.4/9)
Standard deviation 190.4592
You're asked to find:
xbar+1s
xbar-1s
xbar is the mean
For the 1st column
compute xbar-1 time the standard deviation
xbar + 1 time the standard deviation
Then compute xbar+2 times the sd
xbar-2 times the sd; then 3.
Do this for each column separately.
I think 'n' is the number of cases 20.
Find 68 % of 20
95 % of 20 etc.

Can someone help me solve these problems regarding statistics ( Random Variables & probability distribution)?

Can someone help me solve these problems regarding statistics ( Random Variables & probability distribution)

Can someone help me solve these problems regarding statistics ( Random Variables & probability distribution function)

1) The manager of a job shop does not know the probability distribution of the time required to complete an order. However, from past performance she has been able to estimate the mean and variance as 14 days and 2 (days)^2, respectiviely. Find an interval such that the probability is 0.75 that an order is finished during that time?


2) A postal service requires, on the average 2 days to deliver a letter across town. The variance is estimated to be 0.4 (days)^2. If a business executive wants 99 percent of his letters delivered on time, how early should he mail them?


PLEASE provide reasons and steps CLEARLY. For you convenience, here is the answers given in the book.

( In the book it has given answers to the 1st (first) question as k = 2, [14 - 2sqrt(2), 14 + 2sqrt(2)]
and for the 2nd (second) question it has given that k=10 and 2 + k*sqrt(0.4) = 8.3 days)

Please help me solve t\above problems and obtain correct answers.

Thanks.

Can someone help me solve this problem in statistics?

You will want to use the F test for equal variances

the null hypothesis for the F test is:

H0: σ1² / σ2² = 1

The F statistic is 198/225 = 0.88
with 23 and infinite degree of freedom.

p-value = P(F > 0.88) ≈ 0.623692

with a p-value this high you can not reject the null hypothesis. as such we conclude that it is plausible that the variances are equal. We cannot ever conclude that the variances are equal, only that it is plausible.

Note that this test is extremely sensitive to the assumption that the variances come from normal populations. if this is not the case then this test is not valid. even with normal data, this test is not that reliable.

Can someone please help me solve this Statistics problem?

GE, a company that makes light bulbs claims that its bulbs have a mean life
of 800 hours with a standard deviation of 32 hours. Show all work.
a) If you buy a four-pack of bulbs, what is the probability that the mean life
will be 775 to 850 hours for that four-pack?
b) If you buy a case of 40 bulbs, what is the probability that the mean life
will be 775 to 850 hours for the entire case?

Can anyone help me solve this statistics problem?

Given that ^p = the sample proportion of customers who prefer Coca-Cola= 55% = 0.55, n= the number of customers selected in this particular random sample=200, p0= the claimed true value of the population proportion to be tested=50%=0.5 and α= level of significance= the chance of wrongly reject H0 when it is in fact true=5%=0.05.

We wish to test
Ho: p=p0
p=50%
i.e.,p=0.5
versus
Ha: p>p0
p>50%
i.e.,p>0.5

Since n*P0= 200*0.5 = 100 and n*(1-P0)=200*(1-0.5)= 100 are both at least greater than 5, then n is considered to be large and so the sampling distribution of ^p is approximately normally distributed and so it is appropriate to describe and estimate the above probability distribution using the z score in this case.

At α=5%=0.05 level of significance,
the rejection point
= zα
= z0.05
= 1.645

The test static
= z
= (^p-p0)/√[p0*(1-p0)/n]
= (0.55-0.5)/√[0.5*(1-0.5)/200]
= 1.41

In this case, we can only reject Ho in favor of Ha if and only if z>zα; or alternatively, if the p-value- the area under the standard normal curve to the right of z is less than our chosen value of α.

Because z=1.41 is smaller instead of greater than our rejection point zα=z0.05=1.645, then we fail to reject Ho in favor of Ha at the α=0.05 level of significance. Then the market consultant cannot conclude that the proportion of customers who prefer Coca-Cola exceeds 50%. In addition, since the p-value of 0.0793-the area under the standard normal curve to the right of z=1.41 is much greater than instead of less than 0.01, then we don't have any evidence at all to support that Ha is true in against Ho.

Hope this helps.

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