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7nadin3 [17]
4 years ago
12

Monthly water bills for a city have a mean of $108.43 and a standard deviation of $36.98. Find the probability that a randomly s

elected bill will have an amount greater than $173, which the city believes might indicate that someone is wasting water. Would a bill that size be considered unusual?
Probability is 0.04, which is unusual as it is less than 5%
Probability is 0.04, which is unusual as it is not less than 5%
Probability is 0.04, which is usual as it is not less than 5%
Probability is 0.04, which is usual as it is less than 5%
Mathematics
1 answer:
olganol [36]4 years ago
5 0

Answer:

Probability is 0.04, which is unusual as it is less than 5%

Step-by-step explanation:

The event is considered to be unusual if the probability of occurrence of that event is less than 5% or 0.05.

We are given that mean and standard deviation are 108.43 and 36.68.

We have to find P(X>173).

P(X>173)=P((x-mean)/sd>(173-108.43)/36.98)

P(X>173)=P(Z>1.75)

P(X>173)=0.5-0.4599

P(X>173)=0.0401.

As the probability is 0.0401 which is less than 0.05, thus a bill with size $173 will be considered unusual.

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Tamara finds the sum of two number cubes rolled at the same time. The chart below shows all possible sums from the 36 possible c
lutik1710 [3]

Answer:

Tamara should expect the sum of the two cubes to be equal to 5 20 times.

Step-by-step explanation:

The sample space of rolling a number cube are:

S = {1, 2, 3, 4, 5, 6}

If two such cubes are rolled together, then the sum of the two cubes will be 5 for the combinations below:

S₁ = {(1, 4), (2, 3), (3, 2) and (4, 1)}

The total number of outcomes will be, <em>N</em> = 36.

Compute the probability that the sum of rolling two numbered cubes as follows:

P(\text{Sum}=5)=\frac{4}{36}=\frac{1}{9}

Let <em>X</em> = number of time the sum of the two numbers on two cubes is 5.

Two numbered cubes are rolled <em>n</em> = 180 times.

The event of getting a sum of 5 in independent of the other results.

The random variable <em>X</em> follows a Binomial distribution with parameters <em>n</em> = 180 and <em>p</em> = \frac{1}{9}.

The expected value of <em>X</em> is:

E(X)=np

Compute the expected number of times Tamara expects the sum of the two cubes to be equal to 5 as follow:

E(X)=np\\=180\times \frac{1}{9}\\=20

Thus, Tamara should expect the sum of the two cubes to be equal to 5 20 times.

8 0
3 years ago
The College Board SAT college entrance exam consists of three parts: math, writing and critical reading (The World Almanac 2012)
Wittaler [7]

Answer:

Yes, there is a difference between the population mean for the math scores and the population mean for the writing scores.

Test Statistics =   \frac{Dbar - \mu_D}{\frac{s_D}{\sqrt{n} } } follows t_n_-  _1 .

Step-by-step explanation:

We are provided with the sample data showing the math and writing scores for a sample of twelve students who took the SAT ;

Let A = Math Scores ,B = Writing Scores  and D = difference between both

So, \mu_A = Population mean for the math scores

       \mu_B = Population mean for the writing scores

 Let \mu_D = Difference between the population mean for the math scores and the population mean for the writing scores.

            <em>  Null Hypothesis, </em>H_0<em> : </em>\mu_A = \mu_B<em>     or   </em>\mu_D<em> = 0 </em>

<em>      Alternate Hypothesis, </em>H_1<em> : </em>\mu_A \neq  \mu_B<em>      or   </em>\mu_D \neq<em> 0</em>

Hence, Test Statistics used here will be;

            \frac{Dbar - \mu_D}{\frac{s_D}{\sqrt{n} } } follows t_n_-  _1    where, Dbar = Bbar - Abar

                                                               s_D = \sqrt{\frac{\sum D_i^{2}-n*(Dbar)^{2}}{n-1}}

                                                               n = 12

Student        Math scores (A)          Writing scores (B)         D = B - A

     1                      540                            474                                   -66

     2                      432                           380                                    -52  

     3                      528                           463                                    -65

     4                       574                          612                                      38

     5                       448                          420                                    -28

     6                       502                          526                                    24

     7                       480                           430                                     -50

     8                       499                           459                                   -40

     9                       610                            615                                       5

     10                      572                           541                                      -31

     11                       390                           335                                     -55

     12                      593                           613                                       20  

Now Dbar = Bbar - Abar = 489 - 514 = -25

 Bbar = \frac{\sum B_i}{n} = \frac{474+380+463+612+420+526+430+459+615+541+335+613}{12}  = 489

 Abar =  \frac{\sum A_i}{n} = \frac{540+432+528+574+448+502+480+499+610+572+390+593}{12} = 514

 ∑D_i^{2} = 22600     and  s_D = \sqrt{\frac{\sum D_i^{2}-n*(Dbar)^{2}}{n-1}} = \sqrt{\frac{22600 - 12*(-25)^{2} }{12-1} } = 37.05

So, Test statistics =   \frac{Dbar - \mu_D}{\frac{s_D}{\sqrt{n} } } follows t_n_-  _1

                            = \frac{-25 - 0}{\frac{37.05}{\sqrt{12} } } follows t_1_1   = -2.34

<em>Now at 5% level of significance our t table is giving critical values of -2.201 and 2.201 for two tail test. Since our test statistics doesn't fall between these two values as it is less than -2.201 so we have sufficient evidence to reject null hypothesis as our test statistics fall in the rejection region .</em>

Therefore, we conclude that there is a difference between the population mean for the math scores and the population mean for the writing scores.

8 0
3 years ago
Help math pls its a drag and drop
STALIN [3.7K]
A goes in stock price decreased
B goes in stock price increased
C goes in stock price remained constant
D goes in stock price decreased
E goes is stock price increased

Hope this helps!
7 0
3 years ago
Read 2 more answers
Laura makes 10.5% commission on each of her sales. How much does she make when she sells a house for 85,000
kykrilka [37]

Answer:

85000/100*10.5 = $8925

Step-by-step explanation:

N/A

5 0
3 years ago
How to prove this???
swat32
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go to right side now

4( \cos^6 A - \sin^6 A)\\&#10;\Rightarrow 4( \cos^3 A - \sin^3 A)(\cos^3 A + \sin^3 A)

use x^3 - y^3 = (x-y)(x^2 + xy + y^2) and x^3 + y^3 = x^2 - xy + y^2

4( \cos^6 A - \sin^6 A)\\ \Rightarrow 4( \cos^3 A - \sin^3 A)(\cos^3 A + \sin^3 A) \\&#10;\Rightarrow  4(\cos A - \sin A)(\cos^2 A + \cos A \sin A + \sin^2 A) \\&#10;~\quad  \quad\cdot ( \cos A + \sin A)(\cos^2 A - \cos A \sin A + \cos^2 A)

so \sin^2 A + \cos^2 A = 1

4( \cos^6 A - \sin^6 A)\\ \Rightarrow 4(\cos A - \sin A)(\cos^2 A + \cos A \sin A + \sin^2 A) \\ ~\quad \quad\cdot ( \cos A + \sin A)(\cos^2 A - \cos A \sin A + \cos^2 A) \\ \Rightarrow 4(\cos^2 A - \sin^2 A)(1 + \cos A \sin A )(1- \cos A \sin A ) \\ \Rightarrow 4(\cos^2 A - \sin^2 A)(1 - \cos^2 A \sin^2 A )\\ \Rightarrow 4(\cos^2 A - \sin^2 A)(1 - \sin^2 A \cos^2 A ) \\&#10; \Rightarrow Left hand side
4 0
3 years ago
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