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we found that weights of adult green sea urchins are normally distributed with mean of 52 g and standard deviation of 17.2 what is the percent of adult green sea urchins that weigh above 40g?
Would I answer this question using the Z= x-u/standard deviation? Or how would I figure out how many urchins weigh above 40g ?
statistics normal-distribution
asked Nov 21 at 3:46
H.Roper
61
closed as off-topic by Micah, Leucippus, Jean-Claude Arbaut, Tom-Tom, Rebellos Nov 21 at 8:48
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Micah, Leucippus, Jean-Claude Arbaut, Rebellos
If this question can be reworded to fit the rules in the help center, please edit the question.
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we found that weights of adult green sea urchins are normally distributed with mean of 52 g and standard deviation of 17.2 what is the percent of adult green sea urchins that weigh above 40g?
Would I answer this question using the Z= x-u/standard deviation? Or how would I figure out how many urchins weigh above 40g ?
statistics normal-distribution
asked Nov 21 at 3:46
H.Roper
61
closed as off-topic by Micah, Leucippus, Jean-Claude Arbaut, Tom-Tom, Rebellos Nov 21 at 8:48
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Micah, Leucippus, Jean-Claude Arbaut, Rebellos
If this question can be reworded to fit the rules in the help center, please edit the question.
This is hard to read, and it's not clear what you're asking.. You might get more comments if the problem statement, and your work towards the solution are clearer
– Nate 8
Nov 21 at 3:50
ok thanks , I tried to clear it up. Does that make more sense?
– H.Roper
Nov 21 at 3:56
add a comment |
up vote
-2
down vote
favorite
up vote
-2
down vote
favorite
we found that weights of adult green sea urchins are normally distributed with mean of 52 g and standard deviation of 17.2 what is the percent of adult green sea urchins that weigh above 40g?
Would I answer this question using the Z= x-u/standard deviation? Or how would I figure out how many urchins weigh above 40g ?
statistics normal-distribution
asked Nov 21 at 3:46
H.Roper
61
we found that weights of adult green sea urchins are normally distributed with mean of 52 g and standard deviation of 17.2 what is the percent of adult green sea urchins that weigh above 40g?
Would I answer this question using the Z= x-u/standard deviation? Or how would I figure out how many urchins weigh above 40g ?
statistics normal-distribution
statistics normal-distribution
asked Nov 21 at 3:46
H.Roper
61
asked Nov 21 at 3:46
H.Roper
61
edited Nov 21 at 3:55
asked Nov 21 at 3:46
H.Roper
61
asked Nov 21 at 3:46
H.Roper
61
asked Nov 21 at 3:46
H.Roper
61
61
closed as off-topic by Micah, Leucippus, Jean-Claude Arbaut, Tom-Tom, Rebellos Nov 21 at 8:48
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Micah, Leucippus, Jean-Claude Arbaut, Rebellos
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Micah, Leucippus, Jean-Claude Arbaut, Tom-Tom, Rebellos Nov 21 at 8:48
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Micah, Leucippus, Jean-Claude Arbaut, Rebellos
If this question can be reworded to fit the rules in the help center, please edit the question.
This is hard to read, and it's not clear what you're asking.. You might get more comments if the problem statement, and your work towards the solution are clearer
– Nate 8
Nov 21 at 3:50
ok thanks , I tried to clear it up. Does that make more sense?
– H.Roper
Nov 21 at 3:56
add a comment |
This is hard to read, and it's not clear what you're asking.. You might get more comments if the problem statement, and your work towards the solution are clearer
– Nate 8
Nov 21 at 3:50
ok thanks , I tried to clear it up. Does that make more sense?
– H.Roper
Nov 21 at 3:56
This is hard to read, and it's not clear what you're asking.. You might get more comments if the problem statement, and your work towards the solution are clearer
– Nate 8
Nov 21 at 3:50
This is hard to read, and it's not clear what you're asking.. You might get more comments if the problem statement, and your work towards the solution are clearer
– Nate 8
Nov 21 at 3:50
ok thanks , I tried to clear it up. Does that make more sense?
– H.Roper
Nov 21 at 3:56
ok thanks , I tried to clear it up. Does that make more sense?
– H.Roper
Nov 21 at 3:56
add a comment |
1 Answer
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Assuming that the weight of the Sea Urchins is normally distributed with those values:
$W $~$ N(52, 17.2^2)$
Where W is a random variable that corresponds to the weight of a Sea Urchin. Now I'm not a big fan of modelling a weight with a normal distribution (see if you can think why?) but let's continue anyway.
You want to know what percentage of the Sea Urchins have a weight greater than 40 grams? Straight away we know that it must be more than 50% because a normal distribution is symmetric and our mean is 52 grams.
The percentage of the population of Sea Urchins that have weight more than 40 grams is just:
$100$%$ P(W>40)$
Where $P(W>40)$ is the probability that a Sea Urchin weighs more than 40 grams.
Note that:
$P(W>40)$ + $P(W<=40)$ $= 1$
Why is this? Because a Sea Urchin must weigh something, so if we counted all the Sea Urchins that weighed less than 40 grams and all the Sea Urchins that weight more than 40 grams well... We've just counted all the Sea Urchins!
$P(W<=40)$ $= 1 -$ $P(W>40)$
Now we just need to figure out how to calculate $P(W<=40)$.
This is when we use the Z Score formula (as you have mentioned).
$Z $~$ N(0, 1)$
Well this is a much nicer normal distribution (The standard normal distribution). If only there was some way to express the probability we want, $P(W<=40)$, in terms of this nice random variable $Z$.
Well there is!
$P(W<=40)$ = $P(Z<=Z_{score})$
So we just need to find the value that corresponds to $W = 40$ on our Standard Normal distribution. Which is the value $Z_{score}$ used above.
Can you take it from here?
answered Nov 21 at 4:30
KnowsNothing
355
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
Assuming that the weight of the Sea Urchins is normally distributed with those values:
$W $~$ N(52, 17.2^2)$
Where W is a random variable that corresponds to the weight of a Sea Urchin. Now I'm not a big fan of modelling a weight with a normal distribution (see if you can think why?) but let's continue anyway.
You want to know what percentage of the Sea Urchins have a weight greater than 40 grams? Straight away we know that it must be more than 50% because a normal distribution is symmetric and our mean is 52 grams.
The percentage of the population of Sea Urchins that have weight more than 40 grams is just:
$100$%$ P(W>40)$
Where $P(W>40)$ is the probability that a Sea Urchin weighs more than 40 grams.
Note that:
$P(W>40)$ + $P(W<=40)$ $= 1$
Why is this? Because a Sea Urchin must weigh something, so if we counted all the Sea Urchins that weighed less than 40 grams and all the Sea Urchins that weight more than 40 grams well... We've just counted all the Sea Urchins!
$P(W<=40)$ $= 1 -$ $P(W>40)$
Now we just need to figure out how to calculate $P(W<=40)$.
This is when we use the Z Score formula (as you have mentioned).
$Z $~$ N(0, 1)$
Well this is a much nicer normal distribution (The standard normal distribution). If only there was some way to express the probability we want, $P(W<=40)$, in terms of this nice random variable $Z$.
Well there is!
$P(W<=40)$ = $P(Z<=Z_{score})$
So we just need to find the value that corresponds to $W = 40$ on our Standard Normal distribution. Which is the value $Z_{score}$ used above.
Can you take it from here?
answered Nov 21 at 4:30
KnowsNothing
355
add a comment |
up vote
0
down vote
Assuming that the weight of the Sea Urchins is normally distributed with those values:
$W $~$ N(52, 17.2^2)$
Where W is a random variable that corresponds to the weight of a Sea Urchin. Now I'm not a big fan of modelling a weight with a normal distribution (see if you can think why?) but let's continue anyway.
You want to know what percentage of the Sea Urchins have a weight greater than 40 grams? Straight away we know that it must be more than 50% because a normal distribution is symmetric and our mean is 52 grams.
The percentage of the population of Sea Urchins that have weight more than 40 grams is just:
$100$%$ P(W>40)$
Where $P(W>40)$ is the probability that a Sea Urchin weighs more than 40 grams.
Note that:
$P(W>40)$ + $P(W<=40)$ $= 1$
Why is this? Because a Sea Urchin must weigh something, so if we counted all the Sea Urchins that weighed less than 40 grams and all the Sea Urchins that weight more than 40 grams well... We've just counted all the Sea Urchins!
$P(W<=40)$ $= 1 -$ $P(W>40)$
Now we just need to figure out how to calculate $P(W<=40)$.
This is when we use the Z Score formula (as you have mentioned).
$Z $~$ N(0, 1)$
Well this is a much nicer normal distribution (The standard normal distribution). If only there was some way to express the probability we want, $P(W<=40)$, in terms of this nice random variable $Z$.
Well there is!
$P(W<=40)$ = $P(Z<=Z_{score})$
So we just need to find the value that corresponds to $W = 40$ on our Standard Normal distribution. Which is the value $Z_{score}$ used above.
Can you take it from here?
answered Nov 21 at 4:30
KnowsNothing
355
add a comment |
up vote
0
down vote
up vote
0
down vote
Assuming that the weight of the Sea Urchins is normally distributed with those values:
$W $~$ N(52, 17.2^2)$
Where W is a random variable that corresponds to the weight of a Sea Urchin. Now I'm not a big fan of modelling a weight with a normal distribution (see if you can think why?) but let's continue anyway.
You want to know what percentage of the Sea Urchins have a weight greater than 40 grams? Straight away we know that it must be more than 50% because a normal distribution is symmetric and our mean is 52 grams.
The percentage of the population of Sea Urchins that have weight more than 40 grams is just:
$100$%$ P(W>40)$
Where $P(W>40)$ is the probability that a Sea Urchin weighs more than 40 grams.
Note that:
$P(W>40)$ + $P(W<=40)$ $= 1$
Why is this? Because a Sea Urchin must weigh something, so if we counted all the Sea Urchins that weighed less than 40 grams and all the Sea Urchins that weight more than 40 grams well... We've just counted all the Sea Urchins!
$P(W<=40)$ $= 1 -$ $P(W>40)$
Now we just need to figure out how to calculate $P(W<=40)$.
This is when we use the Z Score formula (as you have mentioned).
$Z $~$ N(0, 1)$
Well this is a much nicer normal distribution (The standard normal distribution). If only there was some way to express the probability we want, $P(W<=40)$, in terms of this nice random variable $Z$.
Well there is!
$P(W<=40)$ = $P(Z<=Z_{score})$
So we just need to find the value that corresponds to $W = 40$ on our Standard Normal distribution. Which is the value $Z_{score}$ used above.
Can you take it from here?
answered Nov 21 at 4:30
KnowsNothing
355
Assuming that the weight of the Sea Urchins is normally distributed with those values:
$W $~$ N(52, 17.2^2)$
Where W is a random variable that corresponds to the weight of a Sea Urchin. Now I'm not a big fan of modelling a weight with a normal distribution (see if you can think why?) but let's continue anyway.
You want to know what percentage of the Sea Urchins have a weight greater than 40 grams? Straight away we know that it must be more than 50% because a normal distribution is symmetric and our mean is 52 grams.
The percentage of the population of Sea Urchins that have weight more than 40 grams is just:
$100$%$ P(W>40)$
Where $P(W>40)$ is the probability that a Sea Urchin weighs more than 40 grams.
Note that:
$P(W>40)$ + $P(W<=40)$ $= 1$
Why is this? Because a Sea Urchin must weigh something, so if we counted all the Sea Urchins that weighed less than 40 grams and all the Sea Urchins that weight more than 40 grams well... We've just counted all the Sea Urchins!
$P(W<=40)$ $= 1 -$ $P(W>40)$
Now we just need to figure out how to calculate $P(W<=40)$.
This is when we use the Z Score formula (as you have mentioned).
$Z $~$ N(0, 1)$
Well this is a much nicer normal distribution (The standard normal distribution). If only there was some way to express the probability we want, $P(W<=40)$, in terms of this nice random variable $Z$.
Well there is!
$P(W<=40)$ = $P(Z<=Z_{score})$
So we just need to find the value that corresponds to $W = 40$ on our Standard Normal distribution. Which is the value $Z_{score}$ used above.
Can you take it from here?
answered Nov 21 at 4:30
KnowsNothing
355
answered Nov 21 at 4:30
KnowsNothing
355
answered Nov 21 at 4:30
KnowsNothing
355
answered Nov 21 at 4:30
KnowsNothing
355
355
add a comment |
add a comment |
This is hard to read, and it's not clear what you're asking.. You might get more comments if the problem statement, and your work towards the solution are clearer
– Nate 8
Nov 21 at 3:50
ok thanks , I tried to clear it up. Does that make more sense?
– H.Roper
Nov 21 at 3:56