Normal distribution HELP PLEASE I tried at least 3 times.

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.


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%.

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:
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.

Ohh, okay then I shoudn't round at all. I just tried it using the exact values of each and it worked, thank you.