Suppose the heights of a population of $3,000$ adult penguins are approximately normally distributed with a...
I would like to check myself if following my answer is correct: let us consider following problem:
Suppose the heights of a population of $3,000$ adult penguins are approximately normally distributed with a mean of $65$ centimeters and a standard deviation of $5$ centimeters.
(a) Approximately how many of the adult penguins are between $65$ centimeters and $75$ centimeters
tall?
(b) If an adult penguin is chosen at random from the population, approximately what is the probability that the penguin’s height will be less than $60$ centimeters? Give your answer to the nearest 0.05.
so as i know approximately $68$ or $2/3$ fall in the interval of $[mu-sigma,mu+sigma]$,approximately $96$ fall between $[mu-2*sigma,mu+2*sigma]$
and approximately $99$% fall between
$[mu-3*sigma,mu+3*sigma]$
now we are asked between $75$ and $65$,which is equal
$[mu-2*sigma,mu+2*sigma]$
this range,but in this case it is second half range,in this range it would be half of or $48$%,which means that number of penguins would be $3000*0.48=1440$ penguins would be,am i correct?
on (b), less then $60$ means that below $65-5$ or below $[mu-sigma]$ or $16$ percent would be fall in this interval,am i correct?please help me
normal-distribution
edited Feb 26 '18 at 4:53
Palautot Ka
9201518
asked Aug 2 '13 at 5:40
giorgi
1371213
|
show 7 more comments
I would like to check myself if following my answer is correct: let us consider following problem:
Suppose the heights of a population of $3,000$ adult penguins are approximately normally distributed with a mean of $65$ centimeters and a standard deviation of $5$ centimeters.
(a) Approximately how many of the adult penguins are between $65$ centimeters and $75$ centimeters
tall?
(b) If an adult penguin is chosen at random from the population, approximately what is the probability that the penguin’s height will be less than $60$ centimeters? Give your answer to the nearest 0.05.
so as i know approximately $68$ or $2/3$ fall in the interval of $[mu-sigma,mu+sigma]$,approximately $96$ fall between $[mu-2*sigma,mu+2*sigma]$
and approximately $99$% fall between
$[mu-3*sigma,mu+3*sigma]$
now we are asked between $75$ and $65$,which is equal
$[mu-2*sigma,mu+2*sigma]$
this range,but in this case it is second half range,in this range it would be half of or $48$%,which means that number of penguins would be $3000*0.48=1440$ penguins would be,am i correct?
on (b), less then $60$ means that below $65-5$ or below $[mu-sigma]$ or $16$ percent would be fall in this interval,am i correct?please help me
normal-distribution
edited Feb 26 '18 at 4:53
Palautot Ka
9201518
asked Aug 2 '13 at 5:40
giorgi
1371213
I think everything looks good.
– angryavian
Aug 2 '13 at 5:47
only one thing which i did not understand is that,instead of $0.16$,there is $0.15$ in answers
– giorgi
Aug 2 '13 at 5:49
maybe because it is less then $60$,itself $60$ or $1$ % is not counted?
– giorgi
Aug 2 '13 at 6:02
For the first, $41432$ is closer to what is given by the tables. The $68%$ you used is somewhat imprecise. For the second, they rounded to the nearest $.05$. That is what the question asked for. Your $16%$ is closer to the truth.
– André Nicolas
Aug 2 '13 at 6:03
No, regardless of whether the question is asking for $<60$ or $le 60$, the answer is the same. However, I see the reason: they ask you to round to the nearest 0.05
– angryavian
Aug 2 '13 at 6:03
|
show 7 more comments
I would like to check myself if following my answer is correct: let us consider following problem:
Suppose the heights of a population of $3,000$ adult penguins are approximately normally distributed with a mean of $65$ centimeters and a standard deviation of $5$ centimeters.
(a) Approximately how many of the adult penguins are between $65$ centimeters and $75$ centimeters
tall?
(b) If an adult penguin is chosen at random from the population, approximately what is the probability that the penguin’s height will be less than $60$ centimeters? Give your answer to the nearest 0.05.
so as i know approximately $68$ or $2/3$ fall in the interval of $[mu-sigma,mu+sigma]$,approximately $96$ fall between $[mu-2*sigma,mu+2*sigma]$
and approximately $99$% fall between
$[mu-3*sigma,mu+3*sigma]$
now we are asked between $75$ and $65$,which is equal
$[mu-2*sigma,mu+2*sigma]$
this range,but in this case it is second half range,in this range it would be half of or $48$%,which means that number of penguins would be $3000*0.48=1440$ penguins would be,am i correct?
on (b), less then $60$ means that below $65-5$ or below $[mu-sigma]$ or $16$ percent would be fall in this interval,am i correct?please help me
normal-distribution
edited Feb 26 '18 at 4:53
Palautot Ka
9201518
asked Aug 2 '13 at 5:40
giorgi
1371213
I would like to check myself if following my answer is correct: let us consider following problem:
Suppose the heights of a population of $3,000$ adult penguins are approximately normally distributed with a mean of $65$ centimeters and a standard deviation of $5$ centimeters.
(a) Approximately how many of the adult penguins are between $65$ centimeters and $75$ centimeters
tall?
(b) If an adult penguin is chosen at random from the population, approximately what is the probability that the penguin’s height will be less than $60$ centimeters? Give your answer to the nearest 0.05.
so as i know approximately $68$ or $2/3$ fall in the interval of $[mu-sigma,mu+sigma]$,approximately $96$ fall between $[mu-2*sigma,mu+2*sigma]$
and approximately $99$% fall between
$[mu-3*sigma,mu+3*sigma]$
now we are asked between $75$ and $65$,which is equal
$[mu-2*sigma,mu+2*sigma]$
this range,but in this case it is second half range,in this range it would be half of or $48$%,which means that number of penguins would be $3000*0.48=1440$ penguins would be,am i correct?
on (b), less then $60$ means that below $65-5$ or below $[mu-sigma]$ or $16$ percent would be fall in this interval,am i correct?please help me
normal-distribution
normal-distribution
edited Feb 26 '18 at 4:53
Palautot Ka
9201518
asked Aug 2 '13 at 5:40
giorgi
1371213
edited Feb 26 '18 at 4:53
Palautot Ka
9201518
asked Aug 2 '13 at 5:40
giorgi
1371213
edited Feb 26 '18 at 4:53
Palautot Ka
9201518
edited Feb 26 '18 at 4:53
Palautot Ka
9201518
edited Feb 26 '18 at 4:53
Palautot Ka
9201518
9201518
asked Aug 2 '13 at 5:40
giorgi
1371213
asked Aug 2 '13 at 5:40
giorgi
1371213
asked Aug 2 '13 at 5:40
giorgi
1371213
1371213
I think everything looks good.
– angryavian
Aug 2 '13 at 5:47
only one thing which i did not understand is that,instead of $0.16$,there is $0.15$ in answers
– giorgi
Aug 2 '13 at 5:49
maybe because it is less then $60$,itself $60$ or $1$ % is not counted?
– giorgi
Aug 2 '13 at 6:02
For the first, $41432$ is closer to what is given by the tables. The $68%$ you used is somewhat imprecise. For the second, they rounded to the nearest $.05$. That is what the question asked for. Your $16%$ is closer to the truth.
– André Nicolas
Aug 2 '13 at 6:03
No, regardless of whether the question is asking for $<60$ or $le 60$, the answer is the same. However, I see the reason: they ask you to round to the nearest 0.05
– angryavian
Aug 2 '13 at 6:03
|
show 7 more comments
I think everything looks good.
– angryavian
Aug 2 '13 at 5:47
only one thing which i did not understand is that,instead of $0.16$,there is $0.15$ in answers
– giorgi
Aug 2 '13 at 5:49
maybe because it is less then $60$,itself $60$ or $1$ % is not counted?
– giorgi
Aug 2 '13 at 6:02
For the first, $41432$ is closer to what is given by the tables. The $68%$ you used is somewhat imprecise. For the second, they rounded to the nearest $.05$. That is what the question asked for. Your $16%$ is closer to the truth.
– André Nicolas
Aug 2 '13 at 6:03
No, regardless of whether the question is asking for $<60$ or $le 60$, the answer is the same. However, I see the reason: they ask you to round to the nearest 0.05
– angryavian
Aug 2 '13 at 6:03
I think everything looks good.
– angryavian
Aug 2 '13 at 5:47
I think everything looks good.
– angryavian
Aug 2 '13 at 5:47
only one thing which i did not understand is that,instead of $0.16$,there is $0.15$ in answers
– giorgi
Aug 2 '13 at 5:49
only one thing which i did not understand is that,instead of $0.16$,there is $0.15$ in answers
– giorgi
Aug 2 '13 at 5:49
maybe because it is less then $60$,itself $60$ or $1$ % is not counted?
– giorgi
Aug 2 '13 at 6:02
maybe because it is less then $60$,itself $60$ or $1$ % is not counted?
– giorgi
Aug 2 '13 at 6:02
For the first, $41432$ is closer to what is given by the tables. The $68%$ you used is somewhat imprecise. For the second, they rounded to the nearest $.05$. That is what the question asked for. Your $16%$ is closer to the truth.
– André Nicolas
Aug 2 '13 at 6:03
For the first, $41432$ is closer to what is given by the tables. The $68%$ you used is somewhat imprecise. For the second, they rounded to the nearest $.05$. That is what the question asked for. Your $16%$ is closer to the truth.
– André Nicolas
Aug 2 '13 at 6:03
No, regardless of whether the question is asking for $<60$ or $le 60$, the answer is the same. However, I see the reason: they ask you to round to the nearest 0.05
– angryavian
Aug 2 '13 at 6:03
No, regardless of whether the question is asking for $<60$ or $le 60$, the answer is the same. However, I see the reason: they ask you to round to the nearest 0.05
– angryavian
Aug 2 '13 at 6:03
|
show 7 more comments
1 Answer
1
active
oldest
votes
For rounding 16% to the nearest 0.05 percent you should do the following calculations:
First divide 16% by 0.05: (16%/0.05) = 3.2. Then, round 3.2 to the nearest point = 3.
Now, you should multiply 3 to 0.05 = 0.15.
Good Luck
answered Nov 2 '15 at 12:00
user286183
1
add a comment |
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1 Answer
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1 Answer
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active
oldest
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oldest
votes
For rounding 16% to the nearest 0.05 percent you should do the following calculations:
First divide 16% by 0.05: (16%/0.05) = 3.2. Then, round 3.2 to the nearest point = 3.
Now, you should multiply 3 to 0.05 = 0.15.
Good Luck
answered Nov 2 '15 at 12:00
user286183
1
add a comment |
For rounding 16% to the nearest 0.05 percent you should do the following calculations:
First divide 16% by 0.05: (16%/0.05) = 3.2. Then, round 3.2 to the nearest point = 3.
Now, you should multiply 3 to 0.05 = 0.15.
Good Luck
answered Nov 2 '15 at 12:00
user286183
1
add a comment |
For rounding 16% to the nearest 0.05 percent you should do the following calculations:
First divide 16% by 0.05: (16%/0.05) = 3.2. Then, round 3.2 to the nearest point = 3.
Now, you should multiply 3 to 0.05 = 0.15.
Good Luck
answered Nov 2 '15 at 12:00
user286183
1
For rounding 16% to the nearest 0.05 percent you should do the following calculations:
First divide 16% by 0.05: (16%/0.05) = 3.2. Then, round 3.2 to the nearest point = 3.
Now, you should multiply 3 to 0.05 = 0.15.
Good Luck
answered Nov 2 '15 at 12:00
user286183
1
answered Nov 2 '15 at 12:00
user286183
1
answered Nov 2 '15 at 12:00
user286183
1
answered Nov 2 '15 at 12:00
user286183
1
1
add a comment |
add a comment |
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I think everything looks good.
– angryavian
Aug 2 '13 at 5:47
only one thing which i did not understand is that,instead of $0.16$,there is $0.15$ in answers
– giorgi
Aug 2 '13 at 5:49
maybe because it is less then $60$,itself $60$ or $1$ % is not counted?
– giorgi
Aug 2 '13 at 6:02
For the first, $41432$ is closer to what is given by the tables. The $68%$ you used is somewhat imprecise. For the second, they rounded to the nearest $.05$. That is what the question asked for. Your $16%$ is closer to the truth.
– André Nicolas
Aug 2 '13 at 6:03
No, regardless of whether the question is asking for $<60$ or $le 60$, the answer is the same. However, I see the reason: they ask you to round to the nearest 0.05
– angryavian
Aug 2 '13 at 6:03