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34kurt
3 years ago
12

According to a study, an average of six cell phone thefts is reported in San Francisco per day. Assume that the number of report

ed cell phone thefts per day in the city follow the Poisson distribution. What is the probability that four or five cell phones will be reported stolen tomorrow?
Mathematics
1 answer:
lions [1.4K]3 years ago
8 0

Answer:0.29

Step-by-step explanation:

An average of six cell phone thefts is reported in San Francisco per day. This means our mean value, u = 6

For poisson distribution,

P(x=r) = (e^-u×u^r)/r!

probability that four cell phones will be reported stolen tomorrow=

P(x=4)= (e^-6×6^4)/4!

= (0.00248×1296)/4×3×2×1

= 3.21408/24=

0.13392

P(x=5)= (e^-6×6^5)/5!

= (0.00248×7776)/5×4×3×2×1

= 19.28448/120

= 0.1607

probability that four or five cell phones will be reported stolen tomorrow

= P(x=4) + P(x=5)

= 0.13392 + 0.1607

= 0.294624

Approximately 0.29

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Answer:

−a^2 b^2 + 2a + 2b

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Step-by-step explanation:

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8 0
2 years ago
Amirah's family is driving to her uncle's house. The family travels 383.5 miles between 10:15 am and 4:45 pm. Calculate the fami
ddd [48]

Answer:

The family's average rate of travel for the day was 59.0 MPH.

Step-by-step explanation:

From 10:15AM to 4:45PM, they travelled a distance of 383.5 miles and a total of 6 hours and 30 minutes.

There are 60 minutes in an hour. 30 minutes translates into 1/2 an hour or .5

So, 6.5 hours and a total distance of 383.5 miles.

Velocity (MPH) = Distance (in miles) / Time (in hours)

MPH means Miles Per Hour

Variables:

Velocity (MPH) = x

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Time (in hours) = 6.5

Plug in the variables into the formula

x = 383.5 / 6.5

Divide

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5 0
2 years ago
Read 2 more answers
The local oil changing business is very busy on Saturday mornings and is considering expanding. A national study of similar busi
eimsori [14]

Answer:

t=\frac{4.2-3.6}{\frac{1.4}{\sqrt{16}}}=1.714

Reject the null hypothesis if the observed "t" value is less than -2.131 or higher than 2.131  

Rejection Zone: t_{calculated} or t_{calculated}>2.131

In our case since our calculated value is not on the rejection zone we don't have enough evidence to reject the null hypothesis at 5% of significance.

Step-by-step explanation:

Previous concepts  and data given  

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

\bar X=4.2 represent the sample mean  

s=1.4 represent the sample standard deviation  

n=16 represent the sample selected  

\alpha=0.05 significance level  

State the null and alternative hypotheses.    

We need to conduct a hypothesis in order to check if the mean is different from 3.6, the system of hypothesis would be:    

Null hypothesis:\mu = 3.6    

Alternative hypothesis:\mu \neq 3.6    

If we analyze the size for the sample is < 30 and we know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:    

t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}  (1)    

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".    

Calculate the statistic  

We can replace in formula (1) the info given like this:    

t=\frac{4.2-3.6}{\frac{1.4}{\sqrt{16}}}=1.714

Critical values

On this case since we have a bilateral test we need to critical values. We need to use the t distribution with df=n-1=16-1=15 degrees of freedom. The value for \alpha=0.05 and \alpha/2=0.025 so we need to find on the t distribution with 15 degrees of freedom two values that accumulates 0.025 of the ara on each tail. We can use the following excel codes:

"=T.INV(0.025,15)" "=T.INV(1-0.025,15)"

And we got t_{crit}=\pm 2.131    

So the decision on this case would be:

Reject the null hypothesis if the observed "t" value is less than -2.131 or higher than 2.131  

Rejection Zone: t_{calculated} or t_{calculated}>2.131

Conclusion    

In our case since our calculated value is not on the rejection zone we don't have enough evidence to reject the null hypothesis at 5% of significance.

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Answer:

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Answer:

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Step-by-step explanation:

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The attached image shows the graph for given function.

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