Probability Question. Normally distributed

Click For Summary
SUMMARY

The discussion focuses on the statistical analysis of collagen amounts from a plant extract tested for leukemia treatment, which is normally distributed with a mean (μ) of 71 grams per milliliter and a standard deviation (σ) of 8.5 grams per milliliter. The probabilities calculated include a 0.761 chance of collagen levels exceeding 65 grams per milliliter and a 0.937 chance of levels being below 84 grams per milliliter. The percentage of compounds within one standard deviation of the mean remains unresolved, with participants seeking clarification on the correct application of the normal distribution formula.

PREREQUISITES
  • Understanding of normal distribution concepts
  • Familiarity with statistical notation (μ for mean, σ for standard deviation)
  • Knowledge of probability calculations using Z-scores
  • Experience with statistical equations and their applications
NEXT STEPS
  • Study the calculation of Z-scores in normally distributed data
  • Learn how to apply the empirical rule for normal distributions
  • Explore the use of statistical software for probability calculations
  • Investigate the implications of standard deviation in data analysis
USEFUL FOR

Statisticians, data analysts, researchers in biomedical fields, and anyone involved in statistical modeling of normally distributed data.

mah062
Messages
3
Reaction score
0
1) (1 pt) The extract of a plant native to Taiwan has been tested as a possible treatment for Leukemia. One of the chemical compounds produced from the plant was analyzed for a particular collagen. The collagen amount was found to be normally distributed with μ=71 and σ= 8.5 grams per mililiter.

a) What is the probability that the amount of collagen is greater than 65 grams per mililiter?
Answer: 0.761

b) What is the probability that the amount of collagen is less than 84 grams per mililiter?
Answer: 0.937

c) What percentage of compounds formed from the extract of this plant fall within 1σ of μ(Not sure if this is the right symbol but it's supposed to stand for the mean)?
Answer: ?

2) Relevant equations...
P([(x-μ)/σ]<Z<[(x-μ)/σ])

Normally Distributed...
f(x)=[1/(σ√(2pi))][e^(-0.5[(x-μ)/σ]^2)]



3. The Attempt at a Solution .
I've tried everything I can think of. I used P([(62.5-71)/8.5]<Z<[(79.5-71)/8.5]) but it was wrong. I've attempted it 28 different ways. Can anyone help me? Thanks!
 
Physics news on Phys.org
mah062 said:
1) (1 pt) The extract of a plant native to Taiwan has been tested as a possible treatment for Leukemia. One of the chemical compounds produced from the plant was analyzed for a particular collagen. The collagen amount was found to be normally distributed with μ=71 and σ= 8.5 grams per mililiter.

a) What is the probability that the amount of collagen is greater than 65 grams per mililiter?
Answer: 0.761

b) What is the probability that the amount of collagen is less than 84 grams per mililiter?
Answer: 0.937

c) What percentage of compounds formed from the extract of this plant fall within 1σ of μ(Not sure if this is the right symbol but it's supposed to stand for the mean)?
Answer: ?

2) Relevant equations...
P([(x-μ)/σ]<Z<[(x-μ)/σ])

Normally Distributed...
f(x)=[1/(σ√(2pi))][e^(-0.5[(x-μ)/σ]^2)]



3. The Attempt at a Solution .
I've tried everything I can think of. I used P([(62.5-71)/8.5]<Z<[(79.5-71)/8.5]) but it was wrong. I've attempted it 28 different ways. Can anyone help me? Thanks!

In your expression above, where does the number 62.5 come from? Where does the 79.5 come from? Why do you have two inequalities on Z?

RGV
 

Similar threads

Replies
2
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
1
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 8 ·
Replies
8
Views
4K
Replies
5
Views
2K