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Sagot :
Answer:
a) 0.9959 = 99.59% probability that the mean weight is less than 9.3 ounces
b) 0.0129 = 1.29% probability that the mean weight is more than 9.0 ounces
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 zscore 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 pvalue, 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 of 8.0 ounces and a standard deviation of 1.1 ounces.
This means that [tex]\mu = 8, \sigma = 1.1[/tex]
(a) If 5 potatoes are randomly selected, find the probability that the mean weight is less than 9.3 ounces?
[tex]n = 5[/tex] means that [tex]s = \frac{1.1}{\sqrt{5}} = 0.4919[/tex]
This probability is the pvalue of Z when X = 9.3. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
By the Central Limit Theorem
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{9.3 - 8}{0.4919}[/tex]
[tex]Z = 2.64[/tex]
[tex]Z = 2.64[/tex] has a pvalue of 0.9959
0.9959 = 99.59% probability that the mean weight is less than 9.3 ounces
(b) If 6 potatoes are randomly selected, find the probability that the mean weight is more than 9.0 ounces?
[tex]n = 6[/tex] means that [tex]s = \frac{1.1}{\sqrt{6}} = 0.4491[/tex]
This probability is 1 subtracted by the pvalue of Z when X = 9. So
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{9 - 8}{0.4491}[/tex]
[tex]Z = 2.23[/tex]
[tex]Z = 2.23[/tex] has a pvalue of 0.9871
1 - 0.9871 = 0.0129
0.0129 = 1.29% probability that the mean weight is more than 9.0 ounces
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