Question

The weights of bags of corn produced at a factory are normally distributed with a mean of 28 oz and a standard deviation of 2 oz. Consider a shipment of 800 bags of corn. How many of the bags will weigh 26.5 oz or less?

Answer 1

The teacher says the answer should b 181 bags...how do I get that answer?

Answer 2

Also...I might need help on another math problem after this one. Thanks :)

Answer 3

Alright, so do you have the normal curve for standard deviation to look at, if not I can link

Answer 4

http://www.regentsprep.org/regents/math/algtrig/ats2/normal67.gif

Answer 5

@Benji Okay...what is next?

Answer 6

Alright so you have 800 bags of corn. And 2oz is one standard deviation away from the median. So the median - one deviation = 26 oz, right? We need 26.5 oz though. So looking at the graph if we want only 1.5oz off the median, how many standard deviations is that? It's not a full deviation, but rather 75% of the full 2oz. (Sorry if it's getting confusing) Anyways looking at the graph, if we hop .5 away, that's 27oz, and then half of the next .5 we will be at 26.5 oz.

Answer 7

Now we use the percentages, half of the 15% is 7.5%, and then we add all the others because they want 26.5oz or less.

Answer 8

Wow...I am so lost...

Answer 9

@Benji

Answer 10

Hmm..okay. Well 2oz is the standard deviation of your problem. So 1 standard deviation away from the mean is 28 oz - 2 oz = 26 oz. We need 26.5 oz or less for your problem. So instead of subtracting a full standard deviation or 2oz, we only need to subtract 1.5 oz, which is .75 deviation away.

Answer 11

Suppose \(X\) is a random variable that represents the weight of a bag. You're interested in finding the number of bags from the total \(800\) that fall under \(26.5\text{ oz}\), so what you're really looking for is the proportion of all bags under the given weight. If you know the proportion, then you can find the number by simply multiplying the proportion by \(800\).

To find the proportion is a matter computing the probability \(\mathbb{P}(X\le26.5)\), which is made easy by transforming \(X\), which is normally distributed with mean \(\mu=28\) and standard deviation \(\sigma=2\), into the standard normal distribution \(Z\) with mean \(0\) and standard deviation \(1\):
\[Z=\frac{X-\mu}{\sigma}=\frac{X-28}{2}\implies X=2Z+28\]This means you have
\[\mathbb{P}(X\le26.5)=\mathbb{P}(2Z+28\le26.5)=\mathbb{P}(Z\le-0.75)\approx\color{red}p\]and so the number of bags that fall under the desired weight would be \(800\color{red}p\).

Answer 12

SO... 28 oz - 1.5?

Answer 13

That is how much oz away from the mean 26.5 is. And since 1.5oz is not a full deviation away (since the full stand dev is 2 oz) 1.5oz = .75 whereas 2oz= 1.0

Answer 14

Oh, so then I would subtract 1.0 from 28? how does tht lead me with the answer of 181 bags?

Answer 15

No, you are trying to find the percentage of 800 bags that are 26.5 or less oz in weight. All I was trying to show you thus far is that on the graph I gave you, if you go 1.5 to the left, you will have to add all the percentages so (7.5%+9.2%+4.4%+1.7%+0.5%+0.1%) = ?

Answer 16

not 1.5 to the left sorry, 0.75 to the left

Answer 17

I got 0.234 when I added those percentages.

Answer 18

So.. basically...it is 23.4% in other words..

Answer 19

Yes, so that is that percentage of 800 bags?

Answer 20

uh...yes?

Answer 21

What is the percentage of 800 bags? 0.234*800

Answer 22

Oh okay sorry...I missed that out I will calculate it ...:/

Answer 23

I got 187.2

Answer 24

so...18720% ??? O_O

Answer 25

Nono ha, 187.2 bags is the amount

Answer 26

I am not sure the method your teacher show you in class, this might be a more accurate number than 180 bags

Answer 27

eh...weird...she gave me this answer sheet and she says 181...I have to turn it in today online...she is not available today soo...tht is why I am here

Answer 28

It really needs to b 181 or I won't get credit... :(
Basically need to show work but I have been stuck on it for hours...

Answer 29

Unless... we only need to follow a formula? Not sure though..

Answer 30

I am not sure, you can try to ask @HolsterEmission for help, I do not entirely understand that way

Answer 31

@HolsterEmission

Answer 32

Have you ever used a table like this one before?
http://www.stat.ufl.edu/~athienit/Tables/Ztable.pdf

Answer 33

Well, I have seen my teacher use it.

Answer 34

Okay. Have you ever seen him or her use the table to compute a probability like \(\mathbb{P}(Z\le-0.75)\)?

Answer 35

No not really

Answer 36

Alright, so the table in that link gives you the probability \(\mathbb{P}(Z\le z)\) which corresponds to the area under the curve left of the value \(Z=z\). (Remember, \(Z\) is the random variable, while \(z\) is the actual value of the random variable.)

In your case, \(z=-0.75\). To look up the probability associated with this value, find the row and column that matches your \(z\). Go to the entry in row \(-0.7\) and column \(0.05\), which says the probability for \(z=-0.75\) is about \(\color{red}p=0.2266\)