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Vinvika [58]
3 years ago
5

If I have 3 integers that are greater than -5, and less than or equal to 3. The median is -1, no Mode and range is 5, the mean i

s 0. Give the set of 3 numbers.
Mathematics
1 answer:
Fudgin [204]3 years ago
5 0

Answer:

{-2, -1 , 3}

Step-by-step explanation:

When we have a set like:

{x₁, x₂, x₃}

The mode is the value that appears the most, so if there is no mode, then each value appears just one time.

The median is the middle value, here we know that the median is -1, then we can rewrite the set as:

{x₁, -1 , x₃}

The mean is computed as:

Mean = (x₁ + x₂ + x₃)/3

in this case we know that the mean is 0, then:

0 =  (x₁ + x₂ + x₃)/3

then the numerator must be zero, so:

0 =  (x₁ + x₂ + x₃)

replacing the value of x₂ = -1 we get:

0 = (x₁ - 1 + x₃)

where:

-5 < x₁ < -1 < x₃ ≤ 3

Now we can select the values of x₁ and x₃ such that the sum is equal to zero, and it meets the wanted restrictions.

here we can choose x₃ to be equal to 3 (the maximum allowed value), I do this because I noticed that the other values that are larger than -1 will not work (just with quick math).

then:

0 = x₁ - 1 + 3

Now we can solve this for x₁

0 = x₁ + 2

-2 = x₁

Then the set is:

{-2, -1 , 3}

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The null hypothesis is  H_o : \mu  =  13.4000

The alternative hypothesis is  H_a :  \mu \ne  13.4000

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b

The  99% confidence level is   13.3930  < \mu  < 13.3994

Step-by-step explanation:

From the question we are told that

  The sample size is  n =  10

   The  population mean is  \mu =  13.4 000 \  angstroms

   The level of significance is  \alpha =  0.05

   The  sample data is  

13.3987, 13.3957, 13.3902, 13.4015, 13.4001, 13.3918, 13.3965, 13.3925, 13.3946, and 13.4002

Generally the sample mean is mathematically represented as

        \= x =  \frac{13.3987+ 13.3957\cdots +13.4002 }{10}

=>     \= x = 13.3962

Generally the sample standard deviation  is mathematically represented as

    \sigma = \sqrt{\frac{\sum (x_i - \= x)^2}{n} }

=> \sigma = \sqrt{\frac{ (13.3987 - 13.3962)^2 +  (13.3987 - 13.3962)^2 + \cdots + (13.3987 -13.4002)^2  }{10} }

=>  \sigma =0.0039

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The alternative hypothesis is  H_a :  \mu \ne  13.4000

Generally the test statistics is mathematically represented as

      z  =  \frac{\= x - \mu}{\frac{\sigma}{\sqrt{n} } }

=>    z  =  \frac{13.3962 -  13.4000}{\frac{0.0039}{\sqrt{10} } }

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So

       p-value  = 2* 0.001035

      p-value  = 0.00207

So from the obtained value we see that

     p-value  < \alpha

Hence the null hypothesis is rejected

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Given that the confidence level is  99%  then the level of significance is

    \alpha =  (100 -99)\%

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    Z_{\frac{\alpha }{2} } =  2.58

Generally the margin of error is mathematically represented as  

     E =  Z_{\frac{\alpha }{2} } *  \frac{\sigma}{\sqrt{n} }

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