Suppose that Upper X has a discrete uniform distribution f left-parenthesis x right-parenthesis equals StartLayout left-brace1st Row 1st Column 1 divided by 3, 2nd Column x equals 1,2,3 2nd Row 1st Column 0, 2nd Column otherwise EndLayout A random sample of n equals 33 is selected from this population. Find the probability that the sample mean is greater than 2.1 but less than 2.4. Express the final answer to four decimal places (e. g. 0.9876).

the probability that the sample mean is greater than 2.1 but less than 2.4 is 0.2444

Step-by-step explanation:

Given the data in the question;

x        f(x)         xp(x)               x²p(x)

1         1/3        0.33333        0.33333

2        1/3        0.66667        1.33333

3        1/3        1.00000        3.0000

∑                    2.0000          4.6667

∑(xp(x)) = 2

∑(x²p(x)) = 4.6667

Variance σ² = ∑(x²) - ∑(x)² = 4.6667 - (2)² = 4.6667 - 4 = 0.6667

standard deviation σ = √variance = √0.6667 = 0.8165

Now since, n = 33 which is greater than 30, we can use normal approximation

for normal distribution z score ( x-μ)/σ

mean μ = 2

standard deviation = 0.817

sample size n = 33

standard of error σₓ = σ/√n = 0.817/√33 = 0.1422

so probability will be;

p( 2.1  < X < 2.4 ) = p(( 2.1-2)/0.1422) <  x"-μ/σₓ  <  p(( 2.4-2)/0.1422)

= 0.70 < Z < 2.81    

=  1 - ( 0.703 < Z < 2.812 )

FROM Z-SC0RE TABLE

=  1 - ( 0.25804 + 0.49752 )

= 1 - 0.75556

p( 2.1  < X < 2.4 ) = 0.2444

Therefore,  the probability that the sample mean is greater than 2.1 but less than 2.4 is 0.2444

the direct variation says that:

then the equation is of the form is given by:

            where k is the constant variation.

from the given table:

consider any value of x and m(x)=y

x = 2 and m(x)=y = 3

substitute these value in [1]; we have;

divide both sides by 2 we get;

⇒

check:

put x = -1 and m(x) = -1.5

substitute in the equation:

-1.5 = -1.5                   true

therefore, the function m(x) represents a direct variation

graph of this function as shown below: