The body temperatures of a group of healthy adults have a bell-shaped distribution with a mean of 98.11°F and a standard deviation of 0.69°F. Using the empirical rule, find each approximate percentage below.a. What is the approximate percentage of healthy adults with body temperatures within 2 standard deviations of the mean, or between 96.73°F and 99.49°F?b. What is the approximate percentage of healthy adults with body temperatures between 96.04°F and 100.18°F?a. Approximately % of healthy adults in this group have body temperatures within 2 standard deviations of the mean, or between 96.73°F and 99.49°F.(Type an integer or a decimal. Do not round.) Using the accompanying table of data, blood platelet counts of women have a bell-shaped distribution with a mean of 255.2 and a standard deviation of 65.5. (All units are 1000 cells/uL.) Using Chebyshev's theorem, what isknown about the percentage of women with platelet counts that are within 3 standard deviations of the mean? What are the minimum and maximum possible platelet counts that are within 3 standard deviations of the mean?E Click the icon to view the table of platelet counts.Using Chebyshev's theorem, what is known about the percentage of women with platelet counts that are within 3 standard deviations of the mean?At least % of women have platelet counts within 3 standard deviations of the mean.(Round to the nearest integer as needed.)

Answered: body temperatures of a group of healthy…

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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The distribution of heights in a population of women is approximately normal. Sixteen percent of the women have heights less than 62 inches. About 97.5% of the women have heights less than 71 inches. Use the empirical rule to estimate the mean and standard deviation of the heights in this population. Mean: K inches Standard Deviation: inches
Researchers measured the data speeds for a particular smartphone carrier at 50 airports. The highest speed measured was 72.6 Mbps. The complete list of 50 data speeds has a mean of x=18.29 Mbps and a standard deviation of s=19.75 Mbps. a. What is the difference between carrier's highest data speed and the mean of all 50 data speeds? b. How many standard deviations is that [the difference found in part (a)]? c. Convert the carrier's highest data speed to a z score. d. If we consider data speeds that convert to z scores between −2 and 2 to be neither significantly low nor significantly high, is the carrier's highest data speed significant? Question content area bottom Part 1 a. The difference is 54.3154.31 Mbps. (Type an integer or a decimal. Do not round.) Part 2 b. The difference is enter your response here standard deviations. (Round to two decimal places as needed.)
Researchers measured the data speeds for a particular smartphone carrier at 50 airports. The highest speed measured was 72.5 Mbps. The complete list of 50 data speeds has a mean of x=18.25 Mbps and a standard deviation of s=17.86 Mbps. a. What is the difference between carrier's highest data speed and the mean of all 50 data speeds? b. How many standard deviations is that [the difference found in part (a)]? c. Convert the carrier's highest data speed to a z score. d. If we consider data speeds that convert to z scores between −2 and 2 to be neither significantly low nor significantly high, is the carrier's highest data speed significant?
Use z scores to compare the given values. In a recent awards ceremony, the age of the winner for best actor was 28 and the age of the winner for best actress was 44. For all best actors, the mean age is 42.3 years and the standard deviation is 8.4 years. For all best actresses, the mean age is 33.3 years and the standard deviation is 10.7 years. (All ages are determined at the time of the awards ceremony.) Relative to their genders, who had the more extreme age when winning the award, the actor or the actress? Explain. Since the z score for the actor is z = and the z score for the actress is z=, the had the more extreme age. (Round to two decimal places.)
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Human body temperatures for healthy individuals have approximately a normal distribution with mean 98.25 °F and standard deviation .75 °F. (The past accepted value of 98.6 Fahrenheit was obtained by converting the Celsius value of 370, which is correct to the nearest integer.) a. Find the 90th percentile of the distribution. b. Find the 5th percentile of the distribution. c. What temperature separates the coolest 25% from the others?
Use z scores to compare the given values. In a recent awards ceremony, the age of the winner for best actor was 27 and the age of the winner for best actress were 47. For all best actors, the mean age is 42.7 years and the standard deviation is 7.7 years. For all best actresses, the mean age is 32.9 years and the standard deviation is 11.2 years. (All ages are determined at the time of the awards ceremony.) Relative to their genders, who had the more extreme age when winning the award, the actor or the actress? Explain. Since the z score for the actor is z= and the z score for the actress is z=,the▼actress actor had the more extreme age.
Researchers measured the data speeds for a particular smartphone carrier at 50 airports. The highest speed measured was 71.7 Mbps. The complete list of 50 data speeds has a mean of x = 15.45 Mbps and a standard deviation of s = 21.16 Mbps. a. What is the difference between carrier's highest data speed and the mean of all 50 data speeds? b. How many standard deviations is that [the difference found in part (a)]? c. Convert the carrier's highest data speed to az score. d. If we consider data speeds that convert to z scores between -2 and 2 to be neither significantly low nor significantly high, is the carrier's highest data speed significant? a. The difference is Mbps. (Type an integer or a decimal. Do not round.)
A normally distributed data set has a mean of O and a standard deviation of 2. What is the percent of values between -6.0 and 6.0? (make sure to include % sign in answer.)
Researchers measured the data speeds for a particular smartphone carrier at 50 airports. The highest speed measured was 73.7 Mbps. The complete list of 50 data speeds has a mean of x = 16.05 Mbps and a standard deviation of s = 17.75 Mbps. a. What is the difference between carrier's highest data speed and the mean of all 50 data speeds? b. How many standard deviations is that [the difference found in part (a)]? c. Convert the carrier's highest data speed to a z score. d. If we consider data speeds that convert to z scores between -2 and 2 to be neither significantly low nor significantly high, is the carrier's highest data speed significant? ..... a. The difference is Mbps. (Type an integer or a decimal. Do not round.) b. The difference is standard deviations. (Round to two decimal places as needed.) c. The z score is z = (Round to two decimal places as needed.) d. The carrier's highest data speed is Next MacBook F12 DD F11 DII F10 F9 F8 888 00 F7 F6 F5 F4 F3 F2 F1 & ! @ # $ 7 8 4 5 6…
The average driver spends $39 at the gas station each week with a standard deviation of $11. Assuming that the amount a driver spends on gas follows a normal distribution; find the percentage of drivers who spend: 1. More than $67? 2. Less than $25? 3. Between $25 and $67

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