Asked by StudentD
Students in the psychology department consume an average of 5 cups of coffee per day with a standard deviation of 1.75 cups. The number of cups of coffee consumed is normally distributed.
• What proportion of students consume an amount equal to or less than 6 cups a day?
• How many cups would an individual at the 80th percentile drink?
• What proportion of students consume an amount equal to or less than 6 cups a day?
• How many cups would an individual at the 80th percentile drink?
Answers
Answered by
MathGuru
Use z-scores.
Formula:
z = (x - mean)/sd
In the first part, find z.
z = (6 - 5)/1.75
Calculate, then use a z-table to find the proportion.
In the second part, find z using a z-table for the 80th percentile. Plug into the formula, along with mean and standard deviation. Solve for x.
I hope this will help get you started.
Formula:
z = (x - mean)/sd
In the first part, find z.
z = (6 - 5)/1.75
Calculate, then use a z-table to find the proportion.
In the second part, find z using a z-table for the 80th percentile. Plug into the formula, along with mean and standard deviation. Solve for x.
I hope this will help get you started.
Answered by
StudentD
z = (6 - 5)/1.75 = 0.57
.2157+.50= 0.7157x100=71.57%
i.e 71.57%
In the second part, find z using a z-table for the 80th percentile. Plug into the formula, along with mean and standard deviation. Solve for x.
z = (x - mean)/sd
z (sd)= X-mean
z (sd)+mean= X
0.20= (X-5)/0.57
(0.20 x 0.57)+ 5= 5.11
b)= 5.11?
The closest in the table for 80th percentile is z=0.20 or 0.21 if I am looking at the right place? Is the above correct by any chance?
Thanks
.2157+.50= 0.7157x100=71.57%
i.e 71.57%
In the second part, find z using a z-table for the 80th percentile. Plug into the formula, along with mean and standard deviation. Solve for x.
z = (x - mean)/sd
z (sd)= X-mean
z (sd)+mean= X
0.20= (X-5)/0.57
(0.20 x 0.57)+ 5= 5.11
b)= 5.11?
The closest in the table for 80th percentile is z=0.20 or 0.21 if I am looking at the right place? Is the above correct by any chance?
Thanks
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