Original post by zed963
How do you do 7b?
How do you do 7b?
Since 232 is the mean. You can say that P(Z>w-232/5)=0.3. This is because the whole bit is 1 so the half to the right is 0.5.
0.2 is surrounded by the mean and w, so 0.3 is what is left which is the area to the right. Work this out by using the percentage points table then solve for w. If that makes sense
Original post by Super199
Since 232 is the mean. You can say that P(Z>w-232/5)=0.3. This is because the whole bit is 1 so the half to the right is 0.5.
0.2 is surrounded by the mean and w, so 0.3 is what is left which is the area to the right. Work this out by using the percentage points table then solve for w. If that makes sense
0.2 is surrounded by the mean and w, so 0.3 is what is left which is the area to the right. Work this out by using the percentage points table then solve for w. If that makes sense
w=234.62
then how would you do part c?
Original post by vee_wuvshugs
You've done something wrong there I'm certain.
Original post by zed963
w=234.62
then how would you do part c?
then how would you do part c?
mm not sure.
Original post by zed963
You've done something wrong there I'm certain.
yeah I have. Really shouldn't be working this late
Original post by zed963
w=234.62
then how would you do part c?
then how would you do part c?
use a tree diagram
you know the probability of getting (W<232<w) is 0.2 so the probability of not getting that is 0.8 then do jar 1 then jar 2 then collect all the possible probabilities
exam solutions has a good video on this question
Original post by studentwiz
use a tree diagram
you know the probability of getting (W<232<w) is 0.2 so the probability of not getting that is 0.8 then do jar 1 then jar 2 then collect all the possible probabilities
exam solutions has a good video on this question
you know the probability of getting (W<232<w) is 0.2 so the probability of not getting that is 0.8 then do jar 1 then jar 2 then collect all the possible probabilities
exam solutions has a good video on this question
Link?
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