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Sagot :
Answer:
0.8558 = 85.58% probability that the mean weight of the sample babies would differ from the population mean by less than 52 grams.
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 establishes 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 3215 grams, and a variance of 84,681
This means that [tex]\mu = 3215, \sigma = \sqrt{84681} = 291[/tex]
67 babies are sampled at random from the hospital
This means that [tex]n = 67, s = \frac{291}{\sqrt{67}}[/tex]
What is the probability that the mean weight of the sample babies would differ from the population mean by less than 52 grams?
p-value of Z when X = 3215 + 52 = 3267 subtracted by the p-value of Z when X = 3215 - 52 = 3163. So
X = 3267
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
By the Central Limit Theorem
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{3267 - 3215}{\frac{291}{\sqrt{67}}}[/tex]
[tex]Z = 1.46[/tex]
[tex]Z = 1.46[/tex] has a p-value of 0.9279
X = 3163
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{3163 - 3215}{\frac{291}{\sqrt{67}}}[/tex]
[tex]Z = -1.46[/tex]
[tex]Z = -1.46[/tex] has a p-value of 0.0721
0.9279 - 0.0721 = 0.8558
0.8558 = 85.58% probability that the mean weight of the sample babies would differ from the population mean by less than 52 grams.
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