S1 Normal Distribution - HORRIBLE question

Thank you for looking, I've been struggling quite a bit with this one:

Steel bolts of circular cross section are to be produced.
The lengths of the bolts have to lie between 64.5 and 65.4mm, and their diameters have to be between 5.5 and 6.0mm.

A machine produces these bolts so that their lengths are normally distribtued with a mean of 65.4 and a standard deviation of 0.5mm and their diameters are independently normally distributed about a mean of 5.7mm with standard deviation

.

(a) If 1000 bolts are made, how many bolts will not be within the specified limits for length?

(b) If 50 bolts have a diameter greater than 6.0mm, what is the value of

?

I would really appreciate if someone could break down the steps for me, so I can try doing it myself. I did try it but got a reeeally weird answer ._.

Reply 1

Swayum

a) The probability of how many aren't right is 1 - P(they are right).

So P(they are right) = P(64.5 < x < 65.4)

=

P(\frac{64.5-65.4}{0.5} < z < \frac{65.4 - 65.4}{0.5})

= P(-1.8 < z < 0)

Probablity that z < 0 is a 0.5.

Probability that -1.8 < z = 1 - phi(1.8) = 1 - 0.9641 = 0.0359

So P(-1.8 < z < 0) = 0.5 - 0.0359 = 0.4641 = probability they are right.

So probability they're wrong = 1 - probability they are right = 1 - 0.4641 = 0.5359.

Now I *think* you just do 1000 * 0.5359, which is 535.9 so I think 535 bolts won't be of required length.

I can't be bothered to think about part b, sorry! :p:

Reply 2

Prudence Slutsky

Thank you very much!

I got up to the "= P(-1.8 < z < 0)

Probablity that z < 0 is a 0.5" part myself, but screwed up after. Rep'd :smile:

Reply 3

alex_hk90

I think that this is right:
(b) let random variable Y represent diameters of bolts produced by machine
:undefined:

Y ~ N(5.7, s^2)
P(Y > 6.0) = 50 / 1000 = 0.05
P(Z > (6.0 - 5.7) / s) = P(Z > 0.3 / s) = 0.05
z = 1.6449 (using the small percentage points table)
therefore: 0.3 / s = 1.6449
s = 0.3 / 1.6449
s = 0.182 (to 3 s.f.)