probability with mean and standard deviation

#1
The local hardware store sells screws in 1lb bags. Because the screws aren't identical, the number of screws per bag, although distributed normally,varies with a μ=115, and σ=6.

A carpenter needs a total of 590 screws for a particular project. What is the probability that he will have enough screws if he buys five bags?

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I don't know why I have trouble calculating even the simplest probability problems, but they always confuse me. I've been trying to figure out how to do it but nothing that I get makes sense! I'm frustrated and I need to get the correct answer by tonight.

If anyone could point me in the right direction, I would be eternally grateful!

Thanks.
 

Dragan

Super Moderator
#2
The local hardware store sells screws in 1lb bags. Because the screws aren't identical, the number of screws per bag, although distributed normally,varies with a μ=115, and σ=6.

A carpenter needs a total of 590 screws for a particular project. What is the probability that he will have enough screws if he buys five bags?

--

I don't know why I have trouble calculating even the simplest probability problems, but they always confuse me. I've been trying to figure out how to do it but nothing that I get makes sense! I'm frustrated and I need to get the correct answer by tonight.

If anyone could point me in the right direction, I would be eternally grateful!

Thanks.

The key is that you have to add the population means and variances. That is, if the carpenter buys 5 bags then distriubtion will be normally distributed with Mu=575 and Sigma^2 = 180.

As such, use the usual transformation to find the probability i.e.

Z = (590 - 575) / Sqrt [ 180 ].
 
#3
Thanks, Dragan!

This is what I got:
Z = (590 - 575) / Sqrt [ 180 ]= 1.11

Looked up the z-score of 1.11 in the Unit Normal Table in the back of my text book and the proportion that I got that corresponds to that z-score(i used the proportion in the tail) is .1335.

So, the probability that he will have enough screws if he buys five bags=.1335

Is this correct?
 

Dragan

Super Moderator
#4
Thanks, Dragan!

This is what I got:
Z = (590 - 575) / Sqrt [ 180 ]= 1.11

Looked up the z-score of 1.11 in the Unit Normal Table in the back of my text book and the proportion that I got that corresponds to that z-score(i used the proportion in the tail) is .1335.

So, the probability that he will have enough screws if he buys five bags=.1335

Is this correct?
Yes, on second thought, you're interpretation is the correct one.:)
 
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#5
Please why did you use the variance to calculate the standard deviation and not standard deviation divided by square root of 5. Again when I checked 1.11 against the z table answer is 0.8665 whereas -1.11 is 0.1335 but, the result you got is not minus 1.11.
Please could you clarify. Thank u.


Thanks, Dragan!

This is what I got:
Z = (590 - 575) / Sqrt [ 180 ]= 1.11

Looked up the z-score of 1.11 in the Unit Normal Table in the back of my text book and the proportion that I got that corresponds to that z-score(i used the proportion in the tail) is .1335.

So, the probability that he will have enough screws if he buys five bags=.1335

Is this correct?