Answer:
Hospital A is less likely to record an average birth weight of at least 3400 g.
Explanation:
Applying the Central Limit Theorem and the normal probability distribution.
Normal probability distribution
Problems of normally distributed samples are solved using the z-score formula.
In a set with mean
and standard deviation
, the zscore of a measure X is given by:

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.
Applying the Central Limit Theorem to find the z-score.
![Z = (X - \mu)/(\sigma) = \frac{X - \mu}{(s)/(√(n))]()
Probability of an average birth weight of at least 3400 g?
This probability is 1 subtracted by the pvalue of Z when X = 3400.
Mean is 3100, so

Suppose s is the same for both.
Hospital A:
n = 50. So
![Z = \frac{X - \mu}{(s)/(√(n))]()
![Z = \frac{3400 - 3100}{(s)/(√(50))]()

Hospital B:
n = 10. So
![Z = \frac{X - \mu}{(s)/(√(n))]()
![Z = \frac{3400 - 3100}{(s)/(√(10))]()

Comparasion:

This means that hospital A has the higher z-score.
The higher the z-score, the higher the pvalue.
So, for A, 1 subtracted by the pvalue of Z when give a lower value than the 1 subtracted by the pvalue of Z in b. This means that hospital A is less likely to record an average birth weight of at least 3400 g.