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g Q3 [20 points] Suppose that your average heart rate is 72. The average heart rate in your class is 82 and the variance is 25. a[10 points] What percent of the class has a lower heart rate

User Vany
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3 votes

Answer:

2.28% of the class has a lower heart rate

Explanation:

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean
\mu and standard deviation(which is the square root of the variance)
\sigma, the zscore of a measure X is given by:


Z = (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.

In this problem, we have that:


\mu = 82, \sigma = √(25) = 5

What percent of the class has a lower heart rate

Lower than 72, which is the pvalue of Z when X = 72.


Z = (X - \mu)/(\sigma)


Z = (72 - 82)/(5)


Z = -2


Z = -2 has a pvalue of 0.0228

2.28% of the class has a lower heart rate

User Azhar Khattak
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