A particular fruit's weights are normally distributed, with a mean of 774 grams and a standard deviation of 31 grams.
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Answer:
3.84% probability that it has a low birth weight
Step-by-step explanation:
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.
In this problem, we have that:
If we randomly select a baby, what is the probability that it has a low birth weight?
This is the pvalue of Z when X = 2500. So
has a pvalue of 0.0384
3.84% probability that it has a low birth weight