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Chapter 6: Problem 84

A study of human body temperatures using healthy women showed a mean of \(98.4^{\circ} \mathrm{F}\) and a standard deviation of about \(0.70^{\circ} \mathrm{F}\). Assume the temperatures are approximately Normally distributed. a. Find the percentage of healthy women with temperatures below \(98.6^{\circ} \mathrm{F}\) (this temperature was considered typical for many decades). b. What temperature does a healthy woman have if her temperature is at the 76 th percentile?

Short Answer

Expert verified
a. Approximately 61.41% of healthy women have body temperatures below \(98.6^{\circ} \mathrm{F}\). b. A healthy woman at the 76th percentile has a body temperature of approximately \(98.925^{\circ} \mathrm{F}\).

Step by step solution

01

Calculation of Z-score for part a

A Z-score represents how many standard deviations away from the mean a data point is. For the given temperature of \(98.6^{\circ} \mathrm{F}\), the Z-score is calculated using the formula Z = (X - 渭) / 蟽 where X = temperature, 渭 = mean, 蟽 = standard deviation. Here, X = \(98.6^{\circ} \mathrm{F}\), 渭 = \(98.4^{\circ} \mathrm{F}\), and 蟽 = \(0.70^{\circ} \mathrm{F}\). The Z-score = (98.6 - 98.4) / 0.70 = 0.29.
02

Determine the percentage using the Z-table for part a

After determining the Z-score, use a standard normal distribution (Z) table or calculator to find the area under the curve to the left of the calculated Z-score. For Z = 0.29, the area is approximately 0.6141. This value represents the cumulative probability. Thus the percentage of healthy women with body temperatures below \(98.6^{\circ} \mathrm{F}\) is approximately 61.41%.
03

Calculation of Z-score for part b

This part of the problem requires to find Z-score corresponding to the 76th percentile. Since the percentile represents the area to the left of the Z-score in the standard normal distribution, the Z-table or calculator can be used backwards. From the table, the Z-score for the 76th percentile is approximately 0.75.
04

Calculation of temperature for part b

After the Z-score is found, use the formula X = Z*蟽 + 渭 where 蟽 is standard deviation and 渭 is mean to find the actual body temperature. Here, Z = 0.75, 蟽 = \(0.70^{\circ} \mathrm{F}\), and 渭 = \(98.4^{\circ} \mathrm{F}\). Therefore, X = 0.75*0.70 + 98.4 = \(98.925^{\circ} \mathrm{F}\). Hence, a healthy woman who is at the 76th percentile has a body temperature of approximately \(98.925^{\circ} \mathrm{F}\).

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