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?

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 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.

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

Hospital B:

n = 10. So

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.

yes

step-by-step explanation:

start with a line segment pq that we will copy. mark a point r that will be one endpoint of the new line segment. set the compasses' point on the point p of the line segment to be copied. adjust the compasses' width to the point q. the compasses' width is now equal to the length of the line segment pq.

step-by-step explanation: