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dem82 [27]
3 years ago
7

and are independent Gaussian (Normal) random Variables. has mean 13.9 and variance 5.2; has mean 6.9 and variance 3.8. . (a) Cal

culate
Mathematics
1 answer:
omeli [17]3 years ago
8 0

This question is incomplete, the complete question is;

X and Y are independent Gaussian (Normal) random Variables. X has mean 13.9 and variance 5.2; Y has mean 6.9 and variance 3.8. . (a) Calculate P( W> 10)

Answer:

P( W> 10) is 0.1587

Step-by-step explanation:

Given that;

X ⇒ N( 13.9, 5.2 )

Y ⇒ N( 6.9, 3.8 )

W = X - Y

Therefore

E(W) = E(X) - E(Y)

= 13.9 - 6.9 = 7

Var(W) = Var(X) + Var(Y) -2COV(X.Y)

[ COV(X,Y) = 0 because they are independent]

Var(W) = 5.2 + 3.8 + 0

= 9

Therefore

W ⇒ N( 7, 9 )

so

P( W > 10 )

= 1 - P( W ≤ 10 )

= 1 - P( W-7 /3   ≤   10-7 /3 )

= 1 - P( Z ≤ 1 )           [ Z = W-7 / 3 ⇒ N(0, 1)  ]

from Standard normal distribution table, P( Z ≤ 1 )  = 0.8413

so

1 - P( Z ≤ 1 )  = 1 - 0.8413 = 0.1587

Therefore P( W> 10) is 0.1587

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Give a 98% confidence interval for one population mean, given asample of 28 data points with sample mean 30.0 and sample standar
klio [65]

We have a sample of 28 data points. The sample mean is 30.0 and the sample standard deviation is 2.40. The confidence level required is 98%. Then, we calculate α by:

\begin{gathered} 1-\alpha=0.98 \\ \alpha=0.02 \end{gathered}

The confidence interval for the population mean, given the sample mean μ and the sample standard deviation σ, can be calculated as:

CI(\mu)=\lbrack x-Z_{1-\frac{\alpha}{2}}\cdot\frac{\sigma}{\sqrt[]{n}},x+Z_{1-\frac{\alpha}{2}}\cdot\frac{\sigma}{\sqrt[]{n}}\rbrack

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CI(\mu)=\lbrack30.0-Z_{0.99}\cdot\frac{2.40}{\sqrt[]{28}},30.0+Z_{0.99}\cdot\frac{2.40}{\sqrt[]{28}}\rbrack

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Answer:

Step-by-step explanation:

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Answer:

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