Where am I going wrong? (Standard Deviation)

In summary, the weight of a randomly chosen plastic washer is normally distributed with a mean of 5g. To calculate the standard deviation in grams, we can use the given probability of a washer weighing less than 3g, which corresponds to a z-score of -1.16. Using the symmetry of the normal curve, we can find the z-score that gives us a tail area of 0.877, which is 1.16. We then use the formula x = mean + z*sigma to solve for the standard deviation, which is approximately 1.72g.
  • #1
lando45
84
0
Hi, I was trying to solve this question but my answer is different to the one given by my textbook.

"The weight of a randomly chosen plastic washer is normally distributed with mean 5g. Calculate the standard deviation in grams given that the probability that a randomly chosen washer weighs less than 3g is 0.123."

I said X is equivalent to a normal distribution with mean 5 and variance A.
As X is a continuous distribution, the binomial probability of it being less than 3g can be approximated to the normal distribution being less than 2.5g, so:
P(X<2.5) = 0.123
P(z<(2.5-5)/sqrtA)
let sqrtA = B = standard deviation
P(z<-2.5/B)
1 - P(z<2.5/b) = 0.123
P(z<2.5/B) = 0.877
Referring to statistical tables for the normal distribution I found 2.5/B to equal 1.16
This yields a value of B (standard deviation) to be 2.155
But my answer book says 1.72.

Can anyone see where I'm going wrong?
 
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  • #2
lando45 said:
I said X is equivalent to a normal distribution with mean 5 and variance A.
As X is a continuous distribution, the binomial probability of it being less than 3g can be approximated to the normal distribution being less than 2.5g, so:

I don't understand what you're doing here. Why are you "approximating" by 2.5g? And there's no need to bring in the variance at all.

It's a very straightforward problem. You're given the probability of the washer weighing less than 3g to be 0.123 Let's say the value of 3g corresponds to a z-score of Z. Z is obviously negative since 3g is less than the mean of 5g.

Meaning [tex]\int_{-\infty}^Zf(z)dz = 0.123[/tex]

where f(z) is the normal distribution. Often tables are given for positive z-scores and you're supposed to use the symmetry of the normal curve to figure out the negative part. So we'll use the table from here : http://sweb.cz/business.statistics/normal01.jpg

Look at the tail area shaded in that picture. Now picture the tail area in our problem. Imagine reflecting that tail area about the vertical axis (giving the same tail area, except now it's in the positive z-score region). To get the complementary area corresponding to the illustration in the reference, you have to subtract this area from one (the total area under the curve). With me so far?

In mathematical terms, you want :

[tex]\int_{-\infty}^{-Z}f(z)dz = 1 - 0.123 = 0.877[/tex]

Right, now look at the table in the reference, search for 0.877, and you'll find it under a z-score of 1.16. So we know that (-Z) = 1.16 or Z = -1.16

Any value x on the normal curve is given by x = mean + z*sigma, where z is the z-score and sigma is the std deviation.

So set up the equation :

3 = 5 + (-1.16)*sigma and solve for sigma.
 
  • #3


It seems like you are on the right track, but there may be a few minor errors in your calculations. First, the probability that a randomly chosen washer weighs less than 3g is given as 0.123, not 0.177. So the equation should be P(z<2.5/B) = 0.123. Additionally, when finding the value of B, you should use the z-score of 1.16, not 2.5. So the equation should be 0.123 = P(z<1.16/B). Solving for B, you should get a standard deviation of approximately 1.72, which matches the answer given by your textbook. It's possible that you may have made a small mistake in your calculations or used the incorrect z-score. I would recommend double checking your work and using the z-score of 1.16 to see if you get the correct answer.
 

1. What is standard deviation?

Standard deviation is a measure of how spread out a set of data is from its mean. It is calculated by finding the square root of the variance, which is the average of the squared differences from the mean.

2. Why is standard deviation important?

Standard deviation is important because it provides a way to quantify the variability or consistency of a set of data. A smaller standard deviation indicates that the data points are closer to the mean, while a larger standard deviation indicates that the data points are more spread out.

3. How do I calculate standard deviation?

To calculate standard deviation, you first need to find the mean of the data set. Then, for each data point, subtract the mean and square the result. Next, find the average of these squared differences. Finally, take the square root of the average to find the standard deviation.

4. What does a high or low standard deviation indicate?

A high standard deviation indicates that the data points are spread out from the mean, while a low standard deviation indicates that the data points are clustered around the mean. A high standard deviation can also indicate that there is a lot of variability or inconsistency in the data.

5. How can I use standard deviation to identify errors or mistakes in my data?

If you are expecting your data to be consistent and it has a high standard deviation, this could indicate that there may be errors or mistakes in your data. It is important to check for any outliers or incorrect data points that may be contributing to the high standard deviation. Additionally, comparing the standard deviation of your data to similar data sets can also help identify any potential errors or mistakes.

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