Question

In: Statistics and Probability

Consider the normal curve in the figure to the right, which illustrates the relative frequencies in a distribution of systolic blood pressures for a sample of female students.

Consider the normal curve in the figure to the right, which illustrates the relative frequencies in a distribution of systolic blood pressures for a sample of female students. The distribution has a standard deviation of 15.

a. What is the mean of the distribution? Choose the correct answer below.

A. 90

B. 115

C. 125

D. 105

b. Estimate the percentage of students whose blood pressure is less than 100. Choose the correct answer below.

A. 100%

B. 84%

C. 16%

D. 47%

c. Estimate the percentage of students whose blood pressure is between 110 and 130. Choose the correct answer below.

A. 47%

B. 16%

C. 84%

D. 100%

d. Estimate the percentage of students whose blood pressure is greater than 130. Choose the correct answer below.

A. 100%

B. 84%

C. 16%

D. 47%

Solutions

Expert Solution

a).the mean of the distribution is = 115

[ from the given diagram it is clear that is the center most point of the symmetrical curve,that is 115]

we know that,

the diagram showing the distribution of data :-

b)percentage of students whose blood pressure is less than 100 is = 16 %

[explanation:-

.z score of 100 is:-

from the given diagram it is clear that, percentage of data below 1 sd from the mean = (50-34) = 16]

c)percentage of students whose blood pressure is between 110 and 130 is = 47 %

[explanation:-

z score of 110 is:-

z score of 130 is:-

from the empirical rule, we know that, percentage of data within 1 sd from the mean = 68%

the percentage of data between mean and 1 sd above the mean = 34%

so, the data between -0.33 sd below the mean and 1 sd above the mean be some percentage that is in between 68 % and 34 %....i.e, here 47 %]

d).percentage of students whose blood pressure is greater than 130 is = 16 %

[ explanation:-

z score of 130 is:-

from the given diagram it is clear that, percentage of data above 1 sd from the mean = (50-34) % =16 %]

orchestra answered 1 year ago

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