Respuesta :
Answer:
Hospital A is less likely to record an average birth weight of at least 3400 g.
Step-by-step 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 [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.
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.
Applying the Central Limit Theorem to find the z-score.
[tex]Z = \frac{X - \mu}{\sigma} = \frac{X - \mu}{\frac{s}{\sqrt{n}}[/tex]
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 [tex]\mu = 3100[/tex]
Suppose s is the same for both.
Hospital A:
n = 50. So
[tex]Z = \frac{X - \mu}{\frac{s}{\sqrt{n}}[/tex]
[tex]Z = \frac{3400 - 3100}{\frac{s}{\sqrt{50}}[/tex]
[tex]Z = \frac{300\sqrt{50}}{s}[/tex]
Hospital B:
n = 10. So
[tex]Z = \frac{X - \mu}{\frac{s}{\sqrt{n}}[/tex]
[tex]Z = \frac{3400 - 3100}{\frac{s}{\sqrt{10}}[/tex]
[tex]Z = \frac{300\sqrt{10}}{s}[/tex]
Comparasion:
[tex]\frac{300\sqrt{50}}{s} > \frac{300\sqrt{10}}{s}[/tex]
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.
