1 If we assume that the current cranial capacity in humans o

1) If we assume that the current cranial capacity in humans of an isolated population in South America is approximately normally distributed with a mean of 1400ml and a standard deviation of 125ml, find the probability that a person picked randomly from this population will meet the following criteria for cranial capacity.

a) greater than 1500ml

b) less than 1250ml c) between 1300 and 1550ml

d) Calculate the first quartile or the cranial capacity value that corresponds to less than 25th percentile

e) Calculate the cranial capacity value that corresponds to less than 95% (95th percentile)

Solution

1) If we assume that the current cranial capacity in humans of an isolated population in South America is approximately normally distributed with a mean of 1400ml and a standard deviation of 125ml, find the probability that a person picked randomly from this population will meet the following criteria for cranial capacity.

a) greater than 1500ml

z value of 1500, z=(1500-1400)/125 = 0.80

p( x > 1500) = P( z > 0.80) = 0.2119

b) less than 1250ml

z value of 1250, z=(1250-1400)/125 = -1.2

p( x < 1250) = P( z < -1.2) = 0.1151

c) between 1300 and 1550ml

z value of 1300, z=(1300-1400)/125 = -0.8

z value of 1550, z=(1550-1400)/125 = 1.2

P( 1300<x<1550) = P( -0.8<z<1.2)

P( z <1.2) – P( z < -0.8)

= 0.8849 - 0.2119

= 0.673

d) Calculate the first quartile or the cranial capacity value that corresponds to less than 25th percentile

z value for 25^th percentile =-0.674

x value = mean+z*sd = 1400-0.674*125 = 1315.75ml

e) Calculate the cranial capacity value that corresponds to less than 95% (95th percentile)

z value for 95^th percentile = 1.645

x value = mean+z*sd = 1400+1.645*125 = 1605.625ml