Respuesta :
Answer:
a) The probability is 0.5557 = 55.57%.
b) The probability that a sample of 447 candies will have a mean of 0.8542g or greater is 0.9987 = 99.87%.
c) Yes, because there is a very large probability, of 99.87%, that the amount will be at least the one claimed on the label.
Step-by-step explanation:
To solve this question, we need to understand the normal probability distribution and the central limit theorem.
Normal Probability Distribution:
Problems of normal distributions can be solved using the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.
Central Limit Theorem
The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex].
For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.
Mean weight of 0.8616g and a standard deviation of 0.0518
This means that [tex]\mu = 0.8616, \sigma = 0.0518[/tex]
a) If 1 candy is reandomly selected, find the probability that it weights more than 0.8542g.
This is 1 subtracted by the pvalue of Z when X = 0.8542. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{0.8542 - 0.8616}{0.0518}[/tex]
[tex]Z = -0.14[/tex]
[tex]Z = -0.14[/tex] has a pvalue of 0.4443
1 - 0.4443 = 0.5557
The probability is 0.5557 = 55.57%.
b) If 447 candies are reandomly selected find the probability that their mean weight is at least 0.8542 g.
Sample of 447 means that [tex]n = 447, s = \frac{0.0518}{\sqrt{447}} = 0.00245[/tex]
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
By the Central Limit Theorem
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{0.8542 - 0.8616}{0.0245}[/tex]
[tex]Z = -3.02[/tex]
[tex]Z = -3.02[/tex] has a pvalue of 0.0013
1 - 0.0013 = 0.9987
The probability that a sample of 447 candies will have a mean of 0.8542g or greater is 0.9987 = 99.87%.
c) Given these results does it seem that the candy company is providing consumers with the amount claimed on the label?
Yes, because there is a very large probability, of 99.87%, that the amount will be at least the one claimed on the label.
