Respuesta :
Answer:
You would need to go 1.6 standard errors away from 0.09.
Step-by-step explanation:
We can solve this using the normal distribution, because, by the Central Limit Theorem, the sampling distribution of the sample proportions is approximately normal.
Problems of normally distributed samples are 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.
89% of the sample proportions.
50 - 89/2 = 5.5th percentile
50 + 89/2 = 94.5th percentile
5.5th percentile
X when Z has a pvalue of 0.055. So X when Z = -1.6.
So 1.6 standard error away from 0.09.
94.5th percentile
X when Z has a pvalue of 0.945. So X when Z = 1.6.
So 1.6 standard error away from 0.09.
You would need to go 1.6 standard errors away from 0.09.
The number of standard errors is given by the Zscore, hence, the sample proportion containing 89% of the bad apples would be 1.6 standard deviations from the mean.
- Mean of a Normal distribution = 0.5
- The 89th percentile = 0.89
89th percentile can be interpreted as :
- 0.5 ± (0.89)/2 = 0.5 ± 0.445
The Zscore gives the number of standard deviations a certain score or proportion is from the mean.
Using a normal distribution table :
- P(Z < 0.945) = 1.598 = 1.60
Therefore, about 1.6 standard errors from the mean would contain 89% of the proportion of bad Apples.
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