# Probability

• Jul 18th 2012, 05:53 AM
tomjay
Probability
Length of metal strips produced by a machine are normally distributed with mean length of 150 cm and a standard deviation of 10 cm. Find the probability that the length of a randomly selected strip is
i/ Shorter than 165 cm?

ii/ Longer than 170 cm?

iii/ Between 145 cm and 155 cm?
• Jul 18th 2012, 06:02 AM
emakarov
Re: Probability
Use the standardizing of normal random variables and show your work.
• Jul 19th 2012, 08:28 AM
tomjay
Re: Probability
i think the technical term is head up my arse
• Jul 19th 2012, 11:50 AM
emakarov
Re: Probability
Let X be the length of a random metal strip. As the link says, $P(X < x) = P\left(Z < \frac{x-150}{10}\right)$. Here Z is a standard normal random variable with mean 0 and variance 1. The values of P(Z < z) for various z can be found in tables such as this one.
• Aug 4th 2012, 11:24 AM
tomjay
Re: Probability
yeah just not getting this time to hit the books for a while more i think!
• Aug 6th 2012, 12:22 PM
tomjay
Re: Probability
a) 0.0668 or 6.88% probability
b) 0.0228 or 2.28% probability
c) 0

does this sound right???
• Aug 6th 2012, 01:02 PM
emakarov
Re: Probability
Since the mean is 150, the probability that the length is less than 150 is 50%. And the probability that the length is less than 165 is greater than that, i.e., it should be greater than 50%. What you found in i) is P(X > 165) = 1 - P(X < 165). The answer to ii) is correct. The answer to iii) is clearly wrong because it is quite likely that the length is between 145 and 155. P(145 < X < 155) = P(X < 155) - P(X < 145). You can also use another table here which lists the values for P(0 < Z < x). (It is easy to see that P(0 < Z < x) = 0.5 + P(Z < x).) This table is useful because P(145 < X < 155) = P(-0.5 < Z < 0.5) = 2P(0 < Z < 0.5).
• Aug 7th 2012, 10:50 AM
tomjay
Re: Probability
a) 0.9332 or 93.32%

&

c) 0.383 or 38.3%

any better?
• Aug 7th 2012, 11:39 AM
emakarov
Re: Probability
Yes, that's correct.
• Aug 7th 2012, 12:53 PM
tomjay
Re: Probability