Respuesta :
Answer:
0.3222 = 32.22% probability that the mean weight of the sample babies would differ from the population mean by more than 45 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 normally distributed samples are solved using the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation(which is the square root of the variance) [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.
In this problem, we have that:
[tex]\mu = 3366, \sigma = \sqrt{244036} = 494, n = 118, s = \frac{494}{\sqrt{118}} = 45.48[/tex]
Probability if differs by more than 45 grams?
Less than 3366-45 = 3321 or more than 3366+45 = 3411. Since the normal distribution is symmetric, these probabilities are equal. So we find one of them, and multiply by them.
Less than 3321.
pvalue of Z when X = 3321. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
By the Central Limit Theorem
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{3321 - 3366}{45.48}[/tex]
[tex]Z = -0.99[/tex]
[tex]Z = -0.99[/tex] has a pvalue of 0.1611
2*0.1611 = 0.3222
0.3222 = 32.22% probability that the mean weight of the sample babies would differ from the population mean by more than 45 grams.
