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Dafna1 [17]
3 years ago
7

Consider a t distribution with 7 degrees of freedom. Compute P(-1.29 < t < 1.29). Round your answer to at least three deci

mal places.Consider a t distribution with 18 degrees of freedom. Find the value of c such that P(t<= c) = 0.05. Round your answer to at least three decimal places.
Mathematics
1 answer:
ss7ja [257]3 years ago
3 0

Answer:

a) 0.76197086

b) -1.73406361

Step-by-step explanation:

a)

Consider a t distribution with 7 degrees of freedom. Compute P(-1.29 < t < 1.29)

P(-1.29 < t < 1.29) would be the area under the t distribution curve with 7 degrees of freedom between -1.29 and 1.29, that is in the interval (-1.29, 1.29).

This can be done the old style by looking up in a table or by using the technology with a spreadsheet.

In Excel, the function TDIST(x,n,2) with x>0 gives the area outside the interval (-x, x) of the t distribution with n degrees of freedom.

So TDIST(1.29,7,2) gives the area outside (-1.29, 1.29).

If we subtract this value from 1 we get the desired result

Hence  

P(-1.29 < t < 1.29) = 1 - TDIST(1.29,7,2) = 1 - 0.23802914 = 0.76197086

In OpenOffice Calc, the function is the same replacing “,” with “;”  

That is

P(-1.29 < t < 1.29) = 1 - TDIST(1.29;7;2) = 0.76197086

b)

Consider a t distribution with 18 degrees of freedom. Find the value of c such that P(t≤ c) = 0.05

We are looking for a point c such that the area of the t distribution with 18 degrees of freedom to the left of c is 0.05

In Excel, the inverse function of TDIST is TINV.  

TINV(p*2,n) with p>0 gives the point c such that the area of the t distribution with n degrees of freedom to the right of c is p.  

Since <em>the t distribution is symmetric with respect to 0</em>, -c would be a point such that the area to the left of -c is p.

So we want to compute  in Excel

-TINV(0.05*2,18) = -1.73406361

In OpenOffice Calc  

-TINV(0.05*2;18) = -1.73406361

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A) How many ways can 2 integers from 1,2,...,100 be selected
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Answer with explanation:

→Number of Integers from 1 to 100

                                            =100(50 Odd +50 Even)

→50 Even =2,4,6,8,10,12,14,16,...............................100

→50 Odd=1,3,,5,7,9,..................................99.

→Sum of Two even integers is even.

→Sum of two odd Integers is odd.

→Sum of an Odd and even Integer is Odd.

(a)

Number of ways of Selecting 2 integers from 50 Integers ,so that their sum is even,

   =Selecting 2 Even integers from 50 Even Integers , and Selecting 2 Odd integers from 50 Odd integers ,as Order of arrangement is not Important, ,

        =_{2}^{50}\textrm{C}+_{2}^{50}\textrm{C}\\\\=\frac{50!}{(50-2)!(2!)}+\frac{50!}{(50-2)!(2!)}\\\\=\frac{50!}{48!\times 2!}+\frac{50!}{48!\times 2!}\\\\=\frac{50 \times 49}{2}+\frac{50 \times 49}{2}\\\\=1225+1225\\\\=2450

=4900 ways

(b)

Number of ways of Selecting 2 integers from 100 Integers ,so that their sum is Odd,

   =Selecting 1 even integer from 50 Integers, and 1 Odd integer from 50 Odd integers, as Order of arrangement is not Important,

        =_{1}^{50}\textrm{C}\times _{1}^{50}\textrm{C}\\\\=\frac{50!}{(50-1)!(1!)} \times \frac{50!}{(50-1)!(1!)}\\\\=\frac{50!}{49!\times 1!}\times \frac{50!}{49!\times 1!}\\\\=50\times 50\\\\=2500

=2500 ways

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3 years ago
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