Statistics - Question on Sampling
$begingroup$
Here's the question
The scores, $X_1$ and $X_2$, in papers $1$ and $2$ of an examination are normally distributed with means $24.3$ and $31.2$ respectively and standard deviations $3.5$ and $3.1$ respectively. The final mark for each candidate is found by calculating $2X_1 + 1.5X_2$. Find the probability that a random sample of 8 candidates will have a mean final mark of less than $60$.
This is what I have done so far:
Let $Y = 2X_1 + 1.5X_2$
$E(Y) = E(2X_1 + 1.5X_2)$
$E(Y) = 2E(X_1) + 1.5E(X_2)$
$E(Y) = 2times24.3+1.5times31.2$
$therefore E(Y) = 95.4$
Then,
$Var(Y) = Var(2X_1 + 1.5X_2)$
$Var(Y) = 2^2Var(X_1) + 1.5^2E(X_2)$
$Var(Y) = 2^2times3.5^2+1.5^2times3.1^2$
$therefore Var(Y) = 70.6225$
After that,
$bar Y sim N(95.4, frac{70.6225}{8})$
$P(bar Y < 60)$
$= P(Z < frac{60 - 95.4}{sqrt(frac{70.6225}{8})})$
$= P(Z < -11.9145)$
$approx 0$
However, in my textbook, the answer says 0.9351. So I'm not sure where my mistake occurred.
Can someone tell me what my mistake is? Thanks.
statistics random-variables normal-distribution sampling
asked Dec 11 '18 at 23:38
ianc1339ianc1339
133
$endgroup$
add a comment |
$begingroup$
Here's the question
The scores, $X_1$ and $X_2$, in papers $1$ and $2$ of an examination are normally distributed with means $24.3$ and $31.2$ respectively and standard deviations $3.5$ and $3.1$ respectively. The final mark for each candidate is found by calculating $2X_1 + 1.5X_2$. Find the probability that a random sample of 8 candidates will have a mean final mark of less than $60$.
This is what I have done so far:
Let $Y = 2X_1 + 1.5X_2$
$E(Y) = E(2X_1 + 1.5X_2)$
$E(Y) = 2E(X_1) + 1.5E(X_2)$
$E(Y) = 2times24.3+1.5times31.2$
$therefore E(Y) = 95.4$
Then,
$Var(Y) = Var(2X_1 + 1.5X_2)$
$Var(Y) = 2^2Var(X_1) + 1.5^2E(X_2)$
$Var(Y) = 2^2times3.5^2+1.5^2times3.1^2$
$therefore Var(Y) = 70.6225$
After that,
$bar Y sim N(95.4, frac{70.6225}{8})$
$P(bar Y < 60)$
$= P(Z < frac{60 - 95.4}{sqrt(frac{70.6225}{8})})$
$= P(Z < -11.9145)$
$approx 0$
However, in my textbook, the answer says 0.9351. So I'm not sure where my mistake occurred.
Can someone tell me what my mistake is? Thanks.
statistics random-variables normal-distribution sampling
asked Dec 11 '18 at 23:38
ianc1339ianc1339
133
$endgroup$
$begingroup$
Are $X_1,X_2$ independently distributed?
$endgroup$
– StubbornAtom
Dec 12 '18 at 15:42
$begingroup$
@StubbornAtom Yes
$endgroup$
– ianc1339
Dec 13 '18 at 5:07
add a comment |
$begingroup$
Here's the question
The scores, $X_1$ and $X_2$, in papers $1$ and $2$ of an examination are normally distributed with means $24.3$ and $31.2$ respectively and standard deviations $3.5$ and $3.1$ respectively. The final mark for each candidate is found by calculating $2X_1 + 1.5X_2$. Find the probability that a random sample of 8 candidates will have a mean final mark of less than $60$.
This is what I have done so far:
Let $Y = 2X_1 + 1.5X_2$
$E(Y) = E(2X_1 + 1.5X_2)$
$E(Y) = 2E(X_1) + 1.5E(X_2)$
$E(Y) = 2times24.3+1.5times31.2$
$therefore E(Y) = 95.4$
Then,
$Var(Y) = Var(2X_1 + 1.5X_2)$
$Var(Y) = 2^2Var(X_1) + 1.5^2E(X_2)$
$Var(Y) = 2^2times3.5^2+1.5^2times3.1^2$
$therefore Var(Y) = 70.6225$
After that,
$bar Y sim N(95.4, frac{70.6225}{8})$
$P(bar Y < 60)$
$= P(Z < frac{60 - 95.4}{sqrt(frac{70.6225}{8})})$
$= P(Z < -11.9145)$
$approx 0$
However, in my textbook, the answer says 0.9351. So I'm not sure where my mistake occurred.
Can someone tell me what my mistake is? Thanks.
statistics random-variables normal-distribution sampling
asked Dec 11 '18 at 23:38
ianc1339ianc1339
133
$endgroup$
Here's the question
The scores, $X_1$ and $X_2$, in papers $1$ and $2$ of an examination are normally distributed with means $24.3$ and $31.2$ respectively and standard deviations $3.5$ and $3.1$ respectively. The final mark for each candidate is found by calculating $2X_1 + 1.5X_2$. Find the probability that a random sample of 8 candidates will have a mean final mark of less than $60$.
This is what I have done so far:
Let $Y = 2X_1 + 1.5X_2$
$E(Y) = E(2X_1 + 1.5X_2)$
$E(Y) = 2E(X_1) + 1.5E(X_2)$
$E(Y) = 2times24.3+1.5times31.2$
$therefore E(Y) = 95.4$
Then,
$Var(Y) = Var(2X_1 + 1.5X_2)$
$Var(Y) = 2^2Var(X_1) + 1.5^2E(X_2)$
$Var(Y) = 2^2times3.5^2+1.5^2times3.1^2$
$therefore Var(Y) = 70.6225$
After that,
$bar Y sim N(95.4, frac{70.6225}{8})$
$P(bar Y < 60)$
$= P(Z < frac{60 - 95.4}{sqrt(frac{70.6225}{8})})$
$= P(Z < -11.9145)$
$approx 0$
However, in my textbook, the answer says 0.9351. So I'm not sure where my mistake occurred.
Can someone tell me what my mistake is? Thanks.
statistics random-variables normal-distribution sampling
statistics random-variables normal-distribution sampling
asked Dec 11 '18 at 23:38
ianc1339ianc1339
133
asked Dec 11 '18 at 23:38
ianc1339ianc1339
133
asked Dec 11 '18 at 23:38
ianc1339ianc1339
133
asked Dec 11 '18 at 23:38
ianc1339ianc1339
133
asked Dec 11 '18 at 23:38
ianc1339ianc1339
133
133
$begingroup$
Are $X_1,X_2$ independently distributed?
$endgroup$
– StubbornAtom
Dec 12 '18 at 15:42
$begingroup$
@StubbornAtom Yes
$endgroup$
– ianc1339
Dec 13 '18 at 5:07
add a comment |
$begingroup$
Are $X_1,X_2$ independently distributed?
$endgroup$
– StubbornAtom
Dec 12 '18 at 15:42
$begingroup$
@StubbornAtom Yes
$endgroup$
– ianc1339
Dec 13 '18 at 5:07
$begingroup$
Are $X_1,X_2$ independently distributed?
$endgroup$
– StubbornAtom
Dec 12 '18 at 15:42
$begingroup$
Are $X_1,X_2$ independently distributed?
$endgroup$
– StubbornAtom
Dec 12 '18 at 15:42
$begingroup$
@StubbornAtom Yes
$endgroup$
– ianc1339
Dec 13 '18 at 5:07
$begingroup$
@StubbornAtom Yes
$endgroup$
– ianc1339
Dec 13 '18 at 5:07
add a comment |
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$begingroup$
Are $X_1,X_2$ independently distributed?
$endgroup$
– StubbornAtom
Dec 12 '18 at 15:42
$begingroup$
@StubbornAtom Yes
$endgroup$
– ianc1339
Dec 13 '18 at 5:07