Statistics - Question on Sampling












0












$begingroup$


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







share|cite|improve this question









$endgroup$












  • $begingroup$
    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
















0












$begingroup$


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







share|cite|improve this question









$endgroup$












  • $begingroup$
    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














0












0








0





$begingroup$


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







share|cite|improve this question









$endgroup$




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






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










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


















  • $begingroup$
    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










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