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ehidna [41]
3 years ago
6

The systolic blood pressure of 18-year-old women is a roughly bell-shaped distribution with a mean of 120 mmHg and a standard de

viation of 12 mmHg. What percentage of 18-year-old women have a systolic blood pressure between 96 mmHg and 144 mmHg?
A. Approximately 68%
B. Approximately 99.7%
C. Approximately 99.99%
D. Approximately 95%
Mathematics
1 answer:
saw5 [17]3 years ago
8 0

Answer:

C. approximately 99.99%

Step-by-step explanation:

The distribution bell shape is a normal distribution, with mean 120 mmHg and standard deviation 12 mmHg.

The rank between 96 mmHg and 144 mmHg is wider than the rank between mean plus, minus 3 standard deviation (102 mmHg up to 138 mmHg), and this rank is approximately 99.7 %.Therefore  99.99 % includes more closely values in the rank 96 up to 144 mmHg

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Step-by-step explanation:

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

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

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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:

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Z = \frac{32 - 33}{0.7}

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Step-by-step explanation:

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