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Answer:
The proportion of children in this age range between 70 lbs and 85 lbs is of 0.9306.
Step-by-step explanation:
When the distribution is normal, we use 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.
A study suggested that children between the ages of 6 and 11 in the US have an average weightof 74 lbs, with a standard deviation of 2.7 lbs.
This means that [tex]\mu = 74, \sigma = 2.7[/tex]
What proportion of childrenin this age range between 70 lbs and 85 lbs.
This is the pvalue of Z when X = 85 subtracted by the pvalue of Z when X = 70. So
X = 85
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{85 - 74}{2.7}[/tex]
[tex]Z = 4.07[/tex]
[tex]Z = 4.07[/tex] has a pvalue of 1
X = 70
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{70 - 74}{2.7}[/tex]
[tex]Z = -1.48[/tex]
[tex]Z = -1.48[/tex] has a pvalue of 0.0694
1 - 0.0694 = 0.9306
The proportion of children in this age range between 70 lbs and 85 lbs is of 0.9306.
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