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Helga [31]
3 years ago
5

The average score of 100 students taking a statistics final was 70 with a standard deviation of 7. Assuming a normal distributio

n, what test score separates the top 25% of the students from the lower 75% of students?
Mathematics
1 answer:
san4es73 [151]3 years ago
8 0

Answer:

74.72

Step-by-step explanation:

Given

mean\left ( \mu \right )=70

Standard deviation \left ( \sigma \right )=7

For Top 25%

P\left ( Z

and Z value for 75 % is 0.6744

Thus score which separates the Top 25% of students is

=\mu +\sigma \cdot Z

=70+7\cdot 0.6744=74.7208

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Each front tire on a particular type of vehicle is supposed to be filled to a pressure of 26 psi. Suppose the actual air pressur
Ostrovityanka [42]

Answer:

1) K = 7.895 × 10⁻⁶

2) 0.3024

3)  3.6775 × 10⁻²

4) f(x)= \frac{1}{20} +\frac{3x^{2} }{38000}

5) X and Y are not independent variables

6)

h(x\mid y)  = \frac{38000x^2+38000y^2}{3y^2+19000}

7)  0.54967

8)  25.33 psi

σ = 2.875

Step-by-step explanation:

1) Here we have

f(x, y) =\begin{cases} & \text (x^{2}+y^{2}) \right. 20\leq x\leq 30 & \ 0 \, Otherwise\end{cases}

\int_{x}\int_{y} f(x, y)dydx = 1    

\int_{x}( \right )\int_{y} f(x, y)dy)dx = 1

K\int_{x}( \right )\int_{y}(x^{2} +y^{2})dy)dx = 1

K\int_{x}( (x^{2}y +\frac{y^{3}}{3})_{20}^{30})dx = 1

K\int_{x}( (x^{2}(30-20)) +\frac{30^{3}-20^{3}}{3})_{20}^{30})dx = 1

K\int_{x}( (10x^{2})+\frac{19000}{3})_{20}^{30})dx = 1

K( (10\frac{x^{3}}{3})+\frac{19000}{3}x)_{20}^{30})= 1

K( (10\frac{30^{3}-20^{3}}{3})+\frac{19000}{3}(30-20)))_{20}^{30}) = 1

K =\frac{3}{380000}

2) The probability that both tires are underfilled

P(X≤26,Y≤26) =

\int_{20}^{26} \int_{20}^{26}K(x^{2}+y^{2})dydx

=K\int_{x}( \right )\int_{y}(x^{2} +y^{2})dy)dx

= K\int_{x}( (x^{2}y +\frac{y^{3}}{3})_{20}^{26})dx

K\int_{x}( (x^{2}(26-20)) +\frac{26^{3}-20^{3}}{3})_{20}^{26})dx

K\int_{x}( (6x^{2})+\frac{9576}{3})_{20}^{26})dx

K( (6\frac{x^{3}}{3})+\frac{9576}{3}x)_{20}^{26})

K( (6\frac{26^{3}-20^{3}}{3})+\frac{9576}{3}(26-20)))_{20}^{26})

38304\times K =\frac{3\times38304}{380000}

= 0.3024

That is P(X≤26,Y≤26) = 0.3024

3) The probability that the difference in air pressure between the two tires is at most 2 psi is given by

{20 ≤ x ≤ 30, 20 ≤ y ≤ 30, |  x-y | ≤ 2}

{20 ≤ x ≤ 30, 20 ≤ y ≤ 30, \sqrt{(x-y)^2} ≤ 2}

{20 ≤ x ≤ 30, 20 ≤ y ≤ 30, y ≤ x - 2}

Which gives

20 ≤ x ≤ 22 ::      20 ≤ y ≤ x + 2

22 ≤ x ≤ 28 ::      x - 2 ≤ y ≤ x + 2

28 ≤ x ≤ 30 ::      x - 2 ≤ y ≤ 30

From which we derive probability as

P( |  x-y | ≤2) =  \int_{28}^{30} \int_{x-2}^{30}K(x^{2}+y^{2})dydx +  \int_{20}^{22} \int_{20}^{x+2}K(x^{2}+y^{2})dydx +  \int_{22}^{28} \int_{x-2}^{x+2}K(x^{2}+y^{2})dydx

= K (  \int_{28}^{30} \int_{x-2}^{30}K(x^{2}+y^{2})dydx +  \int_{20}^{22} \int_{20}^{x+2}K(x^{2}+y^{2})dydx +  \int_{22}^{28} \int_{x-2}^{x+2}K(x^{2}+y^{2})dydx)

= K\left [ \left (\frac{14804}{15}  \right )+\left (\frac{8204}{15}  \right ) +\left (\frac{46864}{15}  \right )\right ] = \frac{3}{380000}\times \frac{69872}{15} =\frac{4367}{118750} = 3.6775 × 10⁻²

4) The marginal pressure distribution in the right tire is

f_{x}\left ( x \right )=\int_{y} f(x ,y)dy

=K( \right )\int_{y}(x^{2} +y^{2})dy)

= K( (x^{2}y +\frac{y^{3}}{3})_{20}^{30})

K( (x^{2}(30-20)) +\frac{30^{3}-20^{3}}{3})_{20}^{30})

K(10x^{2}+\frac{19000}{3})}

\frac{3}{38000} (10x^{2}+\frac{19000}{3})}

= \frac{1}{20} +\frac{3x^{2} }{38000}

f(x)= \frac{1}{20} +\frac{3x^{2} }{38000}

5) Here we have

The product of marginal distribution given by

f_x(x) f_y(y) = ( \frac{1}{20} +\frac{3x^{2} }{38000})( \frac{1}{20} +\frac{3y^{2} }{38000}) =\frac{(3x^2+1900)(3y^2+1900)}{1444000000}

≠ f(x,y)

X and Y are not independent variables since the product of the marginal distribution is not joint probability distribution function.

6) Here we have the conditional probability of Y given X = x and the conditional probability of X given that Y = y is given by

h(y\mid x) =\frac{f(x,y))}{f_{X}\left (x  \right )}=  Here we have

 

h(y\mid x) =\frac{x^2+y^2}{\frac{1}{20} +\frac{3x^2}{38000} } = \frac{38000x^2+38000y^2}{3x^2+19000}

Similarly, the the conditional probability of X given that Y = y is given by

h(x\mid y) =\frac{x^2+y^2}{\frac{1}{20} +\frac{3y^2}{38000} } = \frac{38000x^2+38000y^2}{3y^2+19000}

7) Here we have

When the pressure in the left tire is at least 25 psi gives

K\int\limits^{25}_{20}  \frac{38000x^2+38000y^2}{3x^2+19000} {} \, dx

Since x = 22 psi, we have

K\int\limits^{25}_{20}  \frac{38000\cdot 25^2+38000y^2}{3\cdot 25^2+19000} {} \, dx = K \int\limits^{25}_{20}  10.066y^2+6291.39, dx = 57041.942\times \frac{3}{380000}= 0.45033

For P(Y≥25) we have

K \int\limits^{30}_{25}  10.066y^2+6291.39, dx = 69624.72\times \frac{3}{380000} = 0.54967

8) The expected pressure is the conditional mean given by

E(Y\mid x) = K\int\limits^{30}_{20} yh(y \mid x)\, dy

E(Y\mid x) = K\int\limits^{30}_{20} 10.066y^3+6291.39y\, dy = \frac{3}{380000} \times 3208609.27153

= 25.33 psi

The standard deviation is given by

Standard \, deviation =\sqrt{Variance}

Variance = K\int\limits^{30}_{20} [y-E(Y\mid x) ]^2h(y \mid x)\, dy

=K\int\limits^{30}_{20} [y-25.33]^2(10.066y^2+6291.39)\, dy

= \frac{3}{380000} \times 1047259.78 = 8.268

The standard deviation = √8.268 = 2.875.

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