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Answer :
I have a similar problem here with a slightly different given.
Heights of men on a baseball team have a bell shaped distrubtion with a mean of 172cm and a standard deviation of 7cm. Using that is the empirical rule, what is the approximate percentage of the men between the following values?
a) 165 cm and 179cm
b) 151cm and 193cm
The solution is:
a. (165-172)/7 = -1, (179-172)/7 = 1, % by empirical rule = 68%
b. (151-172)/7 = -3, (193-172)/7 = 3, % by empirical rule = 99.7%
I hope that by examining the solution for this problem, it could help you answer your problem on your own.
Heights of men on a baseball team have a bell shaped distrubtion with a mean of 172cm and a standard deviation of 7cm. Using that is the empirical rule, what is the approximate percentage of the men between the following values?
a) 165 cm and 179cm
b) 151cm and 193cm
The solution is:
a. (165-172)/7 = -1, (179-172)/7 = 1, % by empirical rule = 68%
b. (151-172)/7 = -3, (193-172)/7 = 3, % by empirical rule = 99.7%
I hope that by examining the solution for this problem, it could help you answer your problem on your own.
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Answer:
a. 68%
b. 99.7%
Step-by-step explanation:
Bell-shaped distribution means Normal distribution.
For finding the percentage first we have to calculate the value of: [tex]\frac{x-\bar x}{\sigma}[/tex]
If its value is ±1, then using empirical formula percentage = 68%
If its value is ±2, then using empirical formula percentage = 95%
and, If its value is ±3, then using empirical formula percentage = 99.7%
a. [tex]\frac{165 - 173}{8} = -1 \ \ and \ \ \frac{181 - 173}{8} = 1[/tex]
Thus, 68% of data lie within 1 standard deviation of the mean.
b. [tex]\frac{149 - 173}{8} = -3 \ \ and \ \ \frac{197 - 173}{8} = 3[/tex]
Thus, 99.7% of data lie within 3 standard deviation of the mean.
