In UK, the average height of a male is 177 cm, and the average height of a female is 163 cm. If the standard deviation for both is 10 cm, and the distribution of heights for each gender is normal, what proportion of males and females are taller than you?
2 Answers
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Here's the answer, with calculation steps show in R:
stddev <- 10
male.point.avg <- 177
female.point.avg <- 163
my.height <- 193 # a relevant piece of information
my.z.male <- (my.height - male.point.avg) / stddev # Get the z-score
my.z.fem <- (my.height - female.point.avg) / stddev # Get the z-score
paste((1-pnorm(my.z.male)) *100, "% of males are taller than me", sep = "")
[1] "5.4799291699558% of males are taller than me"
paste((1-pnorm(my.z.fem)) *100, "% of males are taller than me", sep = "")
[1] "0.13498980316301% of males are taller than me"
Note that by using 1 - pnorm() I found the cumulative probability associated with the z-score for males and females. The multiplication by 100 was to put the result into a percentage format.
If you wanted to combine males and females you could simply average their 2 average heights, assuming the population is 50% of each gender, or take a weighted average based on some more precise gender ratio.
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0
Assume your height is x Z1=(x-177)/10 Z2=(x-163)/10
Then use table to find probability relevant to Z values
self-studytag. Also, please tell us what you have tried already. $\endgroup$