Normal distribution HELP PLEASE I tried at least 3 times. Watch

JanaALEVEL
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Can someone help me with part ii ?
A machine is set to produce nails of length 10 cm, with standard deviation 0.05 cm. The lengths of the nails are normally distributed.
(i) Find the percentage of nails produced between 9.95 cm and 10.08 cm in length. The machine’s setting is moved by a careless apprentice with the consequence that 16% of the nails are under 5.2 cm in length and 20% are over 5.3 cm.
(ii) Find the new mean and standard deviation.
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here's what I did, my answer is wrong, is the whole method wrong or did I do sth wrong in the process, should I have thought of another way to go about this ?
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ghostwalker
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(Original post by JanaALEVEL)
Can someone help me with part ii ?
A machine is set to produce nails of length 10 cm, with standard deviation 0.05 cm. The lengths of the nails are normally distributed.
(i) Find the percentage of nails produced between 9.95 cm and 10.08 cm in length. The machine’s setting is moved by a careless apprentice with the consequence that 16% of the nails are under 5.2 cm in length and 20% are over 5.3 cm.
(ii) Find the new mean and standard deviation.
Name:  76776598_2588706284549095_2358971591456456704_n.jpg
Views: 10
Size:  113.0 KB

here's what I did, my answer is wrong, is the whole method wrong or did I do sth wrong in the process, should I have thought of another way to go about this ?
Method's fine, but you've made a slip when looking up your tables. E.g. -2.14 comes from 1.6%, not 16%.
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JanaALEVEL
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(Original post by ghostwalker)
Method's fine, but you've made a slip when looking up your tables. E.g. -2.14 comes from 1.6%, not 16%.
AHH yes
Thank you!
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JanaALEVEL
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(Original post by ghostwalker)
Method's fine, but you've made a slip when looking up your tables. E.g. -2.14 comes from 1.6%, not 16%.
hello, sorry to bother you again but can you help me in the last part here ?

A factory produces a very large number of steel bars. The lengths of these bars are normally distributed with 33% of them measuring 20.06 cm or more and 12% of them measuring 20.02 cm or less. Write down two simultaneous equations for the mean and standard deviation of the distribution and solve to find values to 4 significant figures. Hence estimate the proportion of steel bars which measure 20.03 cm or more. The bars are acceptable if they measure between 20.02 cm and 20.08 cm. What percentage are rejected as being outside the acceptable range?

The mean is 20.05
Std deviation is 0.0248
20.03 or more : 0.77944

Here's what I did for the last part:
Name:  75231701_1611778432295319_5225021624767479808_n.jpg
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Size:  100.0 KB the answer In the ms is 22.6%
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ghostwalker
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(Original post by JanaALEVEL)
hello, sorry to bother you again but can you help me in the last part here ?

A factory produces a very large number of steel bars. The lengths of these bars are normally distributed with 33% of them measuring 20.06 cm or more and 12% of them measuring 20.02 cm or less. Write down two simultaneous equations for the mean and standard deviation of the distribution and solve to find values to 4 significant figures. Hence estimate the proportion of steel bars which measure 20.03 cm or more. The bars are acceptable if they measure between 20.02 cm and 20.08 cm. What percentage are rejected as being outside the acceptable range?

The mean is 20.05
Std deviation is 0.0248
20.03 or more : 0.77944

Here's what I did for the last part:
Name:  75231701_1611778432295319_5225021624767479808_n.jpg
Views: 8
Size:  100.0 KB the answer In the ms is 22.6%
Your working is fine, and the problem is down to the accuracy of the figures you're using.

S.D is 0.02477 to 4 sig.fig. You've only quoted it to 3. That may resolve the issue; can't be sure as I used a combination of calculator and 4 fig tables, and got 22.6. But, whether just correcting the s.d. resolves it or not, the problem is the number of sig.fig.s you're using. You may need to use a more accurate value of the mean too.
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JanaALEVEL
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(Original post by ghostwalker)
Your working is fine, and the problem is down to the accuracy of the figures you're using.

S.D is 0.02477 to 4 sig.fig. You've only quoted it to 3. That may resolve the issue; can't be sure as I used a combination of calculator and 4 fig tables, and got 22.6. But, whether just correcting the s.d. resolves it or not, the problem is the number of sig.fig.s you're using. You may need to use a more accurate value of the mean too.
Ohh, okay then I shoudn't round at all. I just tried it using the exact values of each and it worked, thank you.
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ghostwalker
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(Original post by JanaALEVEL)
Ohh, okay then I shoudn't round at all. I just tried it using the exact values of each and it worked, thank you.
Just analysing it a bit further, it's the accuracy of the mean that is causing the issue.

20.08- 20.05 gives us 0.03 (using 4sig.fig.)

20.08-20.0491 gives us 0.0309 (more accurate)

So, we have about a 3% error between the two.

They've clearly used a more exacting value than the 4sig.fig. they've asked for in the question.
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