# Probability question....help!

• Apr 11th 2007, 05:04 AM
cce78
Probability question....help!
Hi all, new to these forums... looking good so far :)

Basically I have a question I'm having trouble answering (I really don't like this probability stuff! The rest of the subject is okay). I'm sure there will be some other textbook exercises I'll need help with too.

Anyway, here it is:

A manufacturer claims that the net weight of a bag of nails will weigh 400 grams (net). The bags have a normal, standard distribution with a standard deviation of 15 grams.

a) Find the probability that a bag selected at random has a weight of 370 grams or less.
b) Find the probability of 40 bags (selected at random) will have a mean weight of 370 grams or less.

• Apr 13th 2007, 02:28 AM
cce78
Can anyone help? :confused:

I'm really stuck :(
• Apr 13th 2007, 06:33 AM
CaptainBlack
Quote:

Originally Posted by cce78
Hi all, new to these forums... looking good so far :)

Basically I have a question I'm having trouble answering (I really don't like this probability stuff! The rest of the subject is okay). I'm sure there will be some other textbook exercises I'll need help with too.

Anyway, here it is:

A manufacturer claims that the net weight of a bag of nails will weigh 400 grams (net). The bags have a normal, standard distribution with a standard deviation of 15 grams.

a) Find the probability that a bag selected at random has a weight of 370 grams or less.

As the weigt of a bag is normaly distributed with mean 400 gm and
SD 15 gm the z score of 370 gm is:

z = (370-400)/15 = -2

Now z has a standard normal distribution, so we look up -2 in a table
of the cumulative standard normal distribution and find:

P(z<= -2) = 0.0228,

so the probability of a bag weighing less than 370 gm is 0.0228 or 2.28%.

Quote:

b) Find the probability of 40 bags (selected at random) will have a mean weight of 370 grams or less.
The mean of a sample drawn from a normal population has an expected
value (mean) equal to the population mean (in this case 400 gm), and
SD equal to the population SD/sqrt(N) where N is the sample size (so in this
case the mean has a SD of 15/sqrt(40) ~=2.37 gm.

So we now proceed as before but with the mean and SD for the sample
mean:

z = (370-400)/2.37 = -12.6

Now z has a standard normal distribution, so we look up -12.6 in a table
of the cumulative standard normal distribution and find:

P(z<= -12.6) = 0.0,

so the probability of the mean of 40 bags weighing less than 370 gm is
essentialy 0.

RonL
• Apr 13th 2007, 11:14 PM
cce78
I have another two questions I'm stuck on (have been given too many questions for homework this week!!!) :(

1. X has a binomial distribution. P = 0.4 and there are 35 trials.
a) Find P(X=20)
b) Find P(X>=20)

c) Find the variance of X

d) Find the expected value of X

e) Approximate P(X>=20) using the normal distribution..

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2. You have applied for an export grant from the Federal Government. There are two types of grants: A and B. You estimate that you have a 70% chance of receiving a type A grant and a 40% chance of receive a type B grant. If you receive a type A grant, there is a 90% chance of receiving a type B grant.

a) What is the likelihood of receiving a type A AND type B grant?

b) What is the likelihood of receiving at least one grant?

c) If you receive a type B grant, what is the probability of receiving a type A grant?