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
and standard deviation(which is the square root of the variance)
, the zscore of a measure X is given by:
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
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
and standard deviation
, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean
and standard deviation
.
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:
![\mu = 3366, \sigma = \sqrt{244036} = 494, n = 118, s = \frac{494}{\sqrt{118}} = 45.48](https://tex.z-dn.net/?f=%5Cmu%20%3D%203366%2C%20%5Csigma%20%3D%20%5Csqrt%7B244036%7D%20%3D%20494%2C%20n%20%3D%20118%2C%20s%20%3D%20%5Cfrac%7B494%7D%7B%5Csqrt%7B118%7D%7D%20%3D%2045.48)
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
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
By the Central Limit Theorem
![Z = \frac{X - \mu}{s}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7Bs%7D)
![Z = \frac{3321 - 3366}{45.48}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7B3321%20-%203366%7D%7B45.48%7D)
![Z = -0.99](https://tex.z-dn.net/?f=Z%20%3D%20-0.99)
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.