Calculate the probability that the sample average of the second sample population exceeds the sample mean of...
A normal population has mean 3 and variance 4. A second normal population has mean.
5 and variance 6. Random samples are taken from both size 40 and 36 stocks.
respectively. Calculate the probability that the sample average of the second sample
population exceeds the sample mean of the first sample by 3 units or more.
population.
My attempt:
I think i need find $P(bar{x}leq 3bar{y})$
I know $bar{x}-bar{y}sim N(u_x+u_y,frac{sigma_x^2}{n_x-1}+frac{sigma_y^2}{n_y-1})$ but here i'm stuck. can someone help me?
probability
asked Nov 19 at 0:49
Bvss12
1,744617
add a comment |
A normal population has mean 3 and variance 4. A second normal population has mean.
5 and variance 6. Random samples are taken from both size 40 and 36 stocks.
respectively. Calculate the probability that the sample average of the second sample
population exceeds the sample mean of the first sample by 3 units or more.
population.
My attempt:
I think i need find $P(bar{x}leq 3bar{y})$
I know $bar{x}-bar{y}sim N(u_x+u_y,frac{sigma_x^2}{n_x-1}+frac{sigma_y^2}{n_y-1})$ but here i'm stuck. can someone help me?
probability
asked Nov 19 at 0:49
Bvss12
1,744617
add a comment |
A normal population has mean 3 and variance 4. A second normal population has mean.
5 and variance 6. Random samples are taken from both size 40 and 36 stocks.
respectively. Calculate the probability that the sample average of the second sample
population exceeds the sample mean of the first sample by 3 units or more.
population.
My attempt:
I think i need find $P(bar{x}leq 3bar{y})$
I know $bar{x}-bar{y}sim N(u_x+u_y,frac{sigma_x^2}{n_x-1}+frac{sigma_y^2}{n_y-1})$ but here i'm stuck. can someone help me?
probability
asked Nov 19 at 0:49
Bvss12
1,744617
A normal population has mean 3 and variance 4. A second normal population has mean.
5 and variance 6. Random samples are taken from both size 40 and 36 stocks.
respectively. Calculate the probability that the sample average of the second sample
population exceeds the sample mean of the first sample by 3 units or more.
population.
My attempt:
I think i need find $P(bar{x}leq 3bar{y})$
I know $bar{x}-bar{y}sim N(u_x+u_y,frac{sigma_x^2}{n_x-1}+frac{sigma_y^2}{n_y-1})$ but here i'm stuck. can someone help me?
probability
probability
asked Nov 19 at 0:49
Bvss12
1,744617
asked Nov 19 at 0:49
Bvss12
1,744617
asked Nov 19 at 0:49
Bvss12
1,744617
asked Nov 19 at 0:49
Bvss12
1,744617
asked Nov 19 at 0:49
Bvss12
1,744617
1,744617
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
You need to find $P(overline{y}geqoverline{x}+3)$, i.e. $P(overline{y}-overline{x}geq 3)$. We have $mathbb{E}(overline{Y}-overline{X})=mathbb{E}(overline{Y})-mathbb{E}(overline{X})=5-3=2$. Assuming independence, we also have $text{Var}(overline{Y}-overline{X})=text{Var}(overline{Y})+text{Var}(overline{X})=6/36+4/40=4/15$. Plug "normalcdf(3,1E99,2,$sqrt{4/15}$)" into your calculator to get the answer.
answered Nov 19 at 1:03
Ben W
1,413513
add a comment |
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1 Answer
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You need to find $P(overline{y}geqoverline{x}+3)$, i.e. $P(overline{y}-overline{x}geq 3)$. We have $mathbb{E}(overline{Y}-overline{X})=mathbb{E}(overline{Y})-mathbb{E}(overline{X})=5-3=2$. Assuming independence, we also have $text{Var}(overline{Y}-overline{X})=text{Var}(overline{Y})+text{Var}(overline{X})=6/36+4/40=4/15$. Plug "normalcdf(3,1E99,2,$sqrt{4/15}$)" into your calculator to get the answer.
answered Nov 19 at 1:03
Ben W
1,413513
add a comment |
You need to find $P(overline{y}geqoverline{x}+3)$, i.e. $P(overline{y}-overline{x}geq 3)$. We have $mathbb{E}(overline{Y}-overline{X})=mathbb{E}(overline{Y})-mathbb{E}(overline{X})=5-3=2$. Assuming independence, we also have $text{Var}(overline{Y}-overline{X})=text{Var}(overline{Y})+text{Var}(overline{X})=6/36+4/40=4/15$. Plug "normalcdf(3,1E99,2,$sqrt{4/15}$)" into your calculator to get the answer.
answered Nov 19 at 1:03
Ben W
1,413513
add a comment |
You need to find $P(overline{y}geqoverline{x}+3)$, i.e. $P(overline{y}-overline{x}geq 3)$. We have $mathbb{E}(overline{Y}-overline{X})=mathbb{E}(overline{Y})-mathbb{E}(overline{X})=5-3=2$. Assuming independence, we also have $text{Var}(overline{Y}-overline{X})=text{Var}(overline{Y})+text{Var}(overline{X})=6/36+4/40=4/15$. Plug "normalcdf(3,1E99,2,$sqrt{4/15}$)" into your calculator to get the answer.
answered Nov 19 at 1:03
Ben W
1,413513
You need to find $P(overline{y}geqoverline{x}+3)$, i.e. $P(overline{y}-overline{x}geq 3)$. We have $mathbb{E}(overline{Y}-overline{X})=mathbb{E}(overline{Y})-mathbb{E}(overline{X})=5-3=2$. Assuming independence, we also have $text{Var}(overline{Y}-overline{X})=text{Var}(overline{Y})+text{Var}(overline{X})=6/36+4/40=4/15$. Plug "normalcdf(3,1E99,2,$sqrt{4/15}$)" into your calculator to get the answer.
answered Nov 19 at 1:03
Ben W
1,413513
answered Nov 19 at 1:03
Ben W
1,413513
answered Nov 19 at 1:03
Ben W
1,413513
answered Nov 19 at 1:03
Ben W
1,413513
1,413513
add a comment |
add a comment |
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