Problem with averages and standard deviation.

In summary, the question given provides information about the average retail price and standard deviation of a pound of grapes in a large city. However, it is unclear what type of distribution is being used for the data. Without this information, it is difficult to determine the price range for at least 15/16 of all pounds of grapes sold in the city. Some possibilities for the distribution could be normal or uniform, but more information is needed to accurately solve the problem.
  • #1
nickar1172
20
0
I had this problem on a quiz and obviously got it wrong, I am confused about how to set this question up and what to solve for:

In a large city, the average retail price of a pound of grapes is $1.79, with a standard deviation of 18 cents. Between what values must be the price of at least 15/16 of all pounds of butter sold in this city?

I posted this question in another forum and somebody told me that it was unsolvable but obviously it can be if I got it wrong...
 
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  • #2
I have moved this thread to our Basic Probability and Statistics subforum as it is a better fit.

The question gives information about grapes, and then asks about butter. Perhaps this is why you were told is it unanswerable.

Assuming we are talking about the same product, are you told anything about the distribution of the data? Are they normally distributed?
 
  • #3
nickar1172 said:
I had this problem on a quiz and obviously got it wrong, I am confused about how to set this question up and what to solve for:

In a large city, the average retail price of a pound of grapes is $1.79, with a standard deviation of 18 cents. Between what values must be the price of at least 15/16 of all pounds of butter sold in this city?

I posted this question in another forum and somebody told me that it was unsolvable but obviously it can be if I got it wrong...

Of course the problem is solvable if you have some type of information abut the p.d.f. of the r.v. ... for example if the price [that we call P...] is uniformely distributed from a to b the its mean value is $\mu = \frac{a + b}{2}$ and the standard deviation is $\sigma = \frac{b-a}{\sqrt{12}}$ and You can esasily find a and b...

Kind regards

$\chi$ $\sigma$
 
  • #4
it was a typo its supposed to be grapes and grapes, still looking for help please
 
  • #5
nickar1172 said:
it was a typo its supposed to be grapes and grapes, still looking for help please

Have you been given any information regarding distribution of the data?
 
  • #6
MarkFL said:
Have you been given any information regarding distribution of the data?

No this was all I was given, what type of distribution is also needed because there is an a) part which stated: using the computing formula, find the standard deviatin of the sample:

2, 6, 4, 7, 3, 1
 
  • #7
nickar1172 said:
No this was all I was given, what type of distribution is also needed because there is an a) part which stated: using the computing formula, find the standard deviatin of the sample:

2, 6, 4, 7, 3, 1

The formulas for sample standard deviation $s$ are:

\(\displaystyle s=\sqrt{\frac{\sum\left(x-\overline{x} \right)^2}{n-1}}\)

\(\displaystyle s=\sqrt{\frac{n\left(\sum x^2 \right)-\left(\sum x \right)^2}{n(n-1)}}\)

Can you proceed?
 
  • #8
I got the standard deviation for that question but what does that have to do with the original question that I posted about the grapes?
 
  • #9
nickar1172 said:
I got the standard deviation for that question but what does that have to do with the original question that I posted about the grapes?

I don't see that the two are related.
 
  • #10
exactly my point taken HOW do you solve the original question...
 
  • #11
Does anyone know how to solve this?

In a large city, the average retail price of a pound of grapes is $1.79, with a standard deviation of 18 cents. Between what values must be the price of at least 15/16 of all pounds of grapes sold in this city?
 
  • #12
nickar1172 said:
Does anyone know how to solve this?

In a large city, the average retail price of a pound of grapes is $1.79, with a standard deviation of 18 cents. Between what values must be the price of at least 15/16 of all pounds of grapes sold in this city?

I might be wrong (stats is not a strong area for me), but I really think you need information regarding the distribution of the data.
 
  • #13
The only way I could see answering this problem would be to assume that $\dfrac{15}{16}$ of the area of a normal curve for price is connected to that amount of sales, but that's a stretch. It's a weird problem and I feel (like others) that something is off.
 
  • #14
nickar1172 said:
Does anyone know how to solve this?

In a large city, the average retail price of a pound of grapes is $1.79, with a standard deviation of 18 cents. Between what values must be the price of at least 15/16 of all pounds of grapes sold in this city?

The question can be answered if You get some hypothesis on the p.d.f. of the price P. It must be P>0, so that a normal distribution probably isn't appropriate. In... http://mathhelpboards.com/basic-probability-statistics-23/problem-averages-standard-deviation-8093.html#post37192

... the case of uniform distribution between a and b is considered and in that case is $\displaystyle \mu= \frac{b + a}{2}$ and $\displaystyle \sigma= \frac{b - a}{\sqrt{12}}$. The data You have permit fo find a and b solving the linear system... $\displaystyle b + a = 3.58$$\displaystyle b - a = .18\ \sqrt{12}\ (1)$

... and You obtain a = 1.48 and b = 2.1. The $\frac{15}{16}$ of prices lie in the interval 1.5 < P < 2.08 ...

Kind regards $\chi$ $\sigma$
 

FAQ: Problem with averages and standard deviation.

What is the difference between mean and average?

The terms mean and average are often used interchangeably, but they have slightly different definitions. The mean is the sum of all values in a dataset divided by the number of values. The average, on the other hand, can refer to either the mean or the median (the middle value in a dataset). Therefore, the mean is a type of average, but not all averages are means.

How is standard deviation calculated?

Standard deviation measures the spread of a dataset. It is calculated by finding the difference between each data point and the mean, squaring those differences, taking the average of the squared differences, and then taking the square root of that average. In other words, it is the square root of the variance.

Why is it important to consider both the average and standard deviation?

The average gives us a measure of central tendency, or the typical value in a dataset. However, it does not tell us how much the data points vary from the average. This is where the standard deviation comes in. It gives us a measure of the spread or variability of the data. By considering both the average and standard deviation, we can gain a better understanding of the dataset and make more accurate interpretations and conclusions.

Can standard deviation be negative?

No, standard deviation cannot be negative. It is always a non-negative value, as it is calculated by taking the square root of the variance, which is always positive. A negative value in the calculation of standard deviation would indicate an error in the data or the calculation.

How is the concept of standard deviation used in statistical analysis?

Standard deviation is a crucial tool in statistical analysis. It is used to assess the spread of data, compare the variability of different datasets, and identify outliers. It is also used in hypothesis testing and confidence interval calculations. A lower standard deviation indicates that the data points are closer to the mean, while a higher standard deviation indicates that the data points are more spread out.

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