Find the total recruiting cost. Advertising costs are $2,975. Interview expenses are $1,320 and $1,560. Hiring expenses are $357
1 answer:
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Answer:
The proportion of infants with birth weights between 125 oz and 140 oz is 0.1359 = 13.59%.
Step-by-step explanation:
When the distribution is normal, we use 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.
In this question, we have that:
The proportion of infants with birth weights between 125 oz and 140 oz is
This is the pvalue of Z when X = 140 subtracted by the pvalue of Z when X = 125. So
X = 140
has a pvalue of 0.9772
X = 125
has a pvalue of 0.8413
0.9772 - 0.8413 = 0.1359
The proportion of infants with birth weights between 125 oz and 140 oz is 0.1359 = 13.59%.