Question

Use StatCrunch to determine the count and percentage of observations falling in each of these intervals by following the instructions listed below or using another appropriate counting method. Properly label and list these counts and percentages in your document.

Start in the “Body Temp” data set. Go to Data  Row Selection  Interactive Tools. In the slider selectors box, click the variable Body Temp into the variable box. Then Click compute.

I got this answer:

68% of the data falls within one standard deviations

68% confidence interval of mean

=98.25+/-0.733

=(97.52, 98.98)

95% of the data falls within two standard deviations

95% confidence interval of mean

=98.25+/-2*0.733

=(96.78, 99.72)

99.7% of the data falls within three standard deviations

99.7% confidence interval of mean

=98.25+/-3*0.733

=(96.05, 100.45)

The box that appears has a slider under the words Body Temp that allows you to create ranges of temperatures that you determined in 4d. Use the slider to obtain the count by looking at the “# rows selected” presented in the first line of the box. Calculate the percentages from the counts you obtained.

I need the answer to this and the rest please!

Do each of the three percentages found in part 4e match to what the Empirical Rule predicts? State your answer in a sentence.

Suppose a new student with a body temperature of 98.6 degrees was recorded. Calculate the z-score of this ‘new’ body temperature and explain in a complete sentence what this z-score indicates.

Answer 1

The empirical calculation are done for each part as follow,

68% of the data falls within one standard deviations

The one standard deviation dispersion is define as,

In terms of z score,

The probability is obtained using the standard normal distribution table for z=1 and z=-1,

68% confidence interval of mean = 98.25+/-0.733 = (97.52, 98.98)

Mean = 98.25

Standard deviation = 0.733,

95% of the data falls within two standard deviations

The two standard deviation dispersion is define as,

In terms of z score,

The probability is obtained using the standard normal distribution table for z=2 and z=-2,

95% confidence interval of mean =98.25+/-2*0.733=(96.78, 99.72)

Mean = 98.25

Standard deviation = 0.733,

99.7% of the data falls within three standard deviations

The three standard deviation dispersion is define as,

In terms of z score,

The probability is obtained using the standard normal distribution table for z=3 and z=-3,

99.7% confidence interval of mean =98.25+/-3*0.733=(96.05,100.45)

Mean = 98.25

Standard deviation = 0.733,

Z-score for student with body temperature of 98.6 degrees

The z score is obtained using the formula,

The z score indicates how many standard deviations the value is from the mean. Here we can see that the body temperature 98.6 degree is 0.478 standard deviation away from the mean.