Here is a random sample of the body temperature of 25 young adults. 96 96.6 96.7...

Here is a random sample of the body temperature of 25 young adults.

96	96.6	96.7	96.9	97
97.1	97.1	97.2	97.3	97.4
97.4	97.7	97.7	97.7	97.8
97.9	98	98	98.2	98.2
98.3	98.3	98.7	98.8	98.9

Complete the relative frequency distribution table.

Temperature Group	Frequency	Relative Frequency
96 ≤ x < 96.41	1	1/25
96.41 ≤ x < 96.82	2	2/25
96.82 ≤ x < 97.23	5	5/25
97.23 ≤ x < 97.64	3	3/25
97.64 ≤ x < 98.05	7	7/25
98.05 ≤ x < 98.46	4	4/25
98.46 ≤ x < 98.87	2	2/25
98.87 ≤ x < 99.28	1	1/25

Expert Solution

Steps to do the problem

1. Arrange the dataset in ascending order.

2. Look at the first interval and check how many points in the dataset lies in that interval. Note that number under frequency.

3. Relative frequency = Frequency in a class / Divide by total number of data points.

4. Cumulative frequency = Add each frequency one after an other as shown below

milcah answered 1 year ago

The following data are for pre-exercise and post-exercise body temperature measure- ments for eight young adults....

The following data are for pre-exercise and post-exercise body temperature measure- ments for eight young adults. Individual Number 1 2 3 4 5 6 7 8 Resting (R) 99.0 97.8 98.6 98.7 98.7 98.2 98.7 98.6 Post-Exercise (P) 99.4 98.1 98.6 98.7 98.7 98.2 98.8 99.2 (a) In R, Define a vector (b) In R, Define a vector (c) In R, Define a vector R for the pre-exercise (i.e., resting) body temperature. P for the pre-exercise (i.e., resting) body temperature....

Mimi conducted a survey on a random sample of 100 adults. 58 adults in the sample...

Mimi conducted a survey on a random sample of 100 adults. 58 adults in the sample chose banana as his / her favorite fruit. Construct a 95% confidence interval estimate of the proportion of adults whose favorite fruit is banana. Show all work. Just the answer, without supporting work, will receive no credit.

1. A sample of 96 observations showed an average of 25 and a standard deviation of...

1. A sample of 96 observations showed an average of 25 and a standard deviation of 16. Construct the 90% confidence interval for the population mean. What is the lower limit of this interval? NOTE: WRITE YOUR ANSWER USING 4 DECIMAL DIGITS. DO NOT ROUND UP OR DOWN. 2. In a recent poll, from a sample of 140 likely voters, 65 answered they approve the way the president is handling his job. Compute the upper limit of the 95% confidence...

1. Human Body Temperature. A sample of 115 body temperatures with a mean of 98.20℉ and...

1. Human Body Temperature. A sample of 115 body temperatures with a mean of 98.20℉ and a standard deviation of 0.62℉. Use a 0.05 significance level to test the claim that the mean body temperature of the population is equal to 97.6℉, as is commonly believed. Is there sufficient evidence to conclude that the common belief wrong? 2. Smartphone Users. A Ring Central survey of 395 smartphone users showed that 158 of them said that their smartphone is the only...

The body temperature of adults is normally distributed with a mean of 98.6°F and standard deviation...

The body temperature of adults is normally distributed with a mean of 98.6°F and standard deviation of 0.60°F. Suppose 36 adults are randomly selected, what is the probability that their mean body temperature is less than 98.0°F?

Physical activity of obese young adults. In a study on the physical activity of young adults,...

Physical activity of obese young adults. In a study on the physical activity of young adults, pediatric researchers measured overall physical activity as the total number of registered movements (counts) over a period of time and then computed the number of counts per minute (cpm) for each subject (International Journal of Obesity, Jan. 2007). The study revealed that the overall physical activity of obese young adults has a mean of μ = 320 cpm μ = 320 cpm and a...

A researcher is wondering whether the smoking habits of young adults (18-25 years of age) in...

A researcher is wondering whether the smoking habits of young adults (18-25 years of age) in a certain city in the U.S. are the same as the proportion of the general population of young adults in the U.S. A recent study stated that the proportion of young adults who reported smoking at least twice a week or more in the last month was 0.16. The researcher collected data from a random sample of 75 adults in the city of interest,...

2) Human Body Temperature. A sample of 112 body temperatures with a mean of 98.20°F and...

2) Human Body Temperature. A sample of 112 body temperatures with a mean of 98.20°F and a standard deviation of 0.62°F. Use a 0.05 significance level to test the claim that the mean body temperature of the population is equal to 99.6°F, as is commonly believed. Is there sufficient evidence to conclude that the common belief wrong?

According to a recent study, 22.6% of adults are smokers. A random sample of 300 adults...

According to a recent study, 22.6% of adults are smokers. A random sample of 300 adults is obtained. Describe the sampling distribution of the sample proportion of adults who smoke. In a random sample of 300 adults, what is the probability that at least 50 are smokers? Would it be unusual if a random sample of 300 adults resulted in 17% or less being smokers?

In an opinion survey, a random sample of 1000 adults from the U.S.A. and 1000 adults...

In an opinion survey, a random sample of 1000 adults from the U.S.A. and 1000 adults from Germany were asked whether they supported the death penalty. 560 American adults and 590 German adults indicated that they supported the death penalty. Researcher wants to know if there is sufficient evidence to conclude that the proportion of adults who support the deal penally in the US is different than that in Germany. Use a confidence interval (at confidence interval of 98%) to...

Question