Mathematics, 05.06.2020 02:59 01030901

Suppose that the mean birth weight of human babies is 3100 g. Hospital A records an average of 50 births a day. Hospital B records an average of 10 births a day. On a particular day, which hospital is less likely to record an average birth weight of at least 3400 g?

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Answer from: rishawn12345

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 $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

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 $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

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.

$Z = \frac{X - \mu}{\sigma} = \frac{X - \mu}{\frac{s}{\sqrt{n}}$

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 $\mu = 3100$

Suppose s is the same for both.

Hospital A:

n = 50. So

$Z = \frac{X - \mu}{\frac{s}{\sqrt{n}}$

$Z = \frac{3400 - 3100}{\frac{s}{\sqrt{50}}$

$Z = \frac{300\sqrt{50}}{s}$

Hospital B:

n = 10. So

$Z = \frac{X - \mu}{\frac{s}{\sqrt{n}}$

$Z = \frac{3400 - 3100}{\frac{s}{\sqrt{10}}$

$Z = \frac{300\sqrt{10}}{s}$

Comparasion:

$\frac{300\sqrt{50}}{s} \frac{300\sqrt{10}}{s}$

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.

Answer from: Quest

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Answer from: Quest

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step-by-step explanation:

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