Assume that the heights of females in a certain village are

Assume that the heights of females in a certain village are normally distributed with a mean of 66 in. and standard deviation of 4 in. What percent of these females are between 74 in. and 78 in. tall? Areas Under the StandardNormal Curve (the z-table) z .00 1.00 2.00 3.00 A .000 .341 .477 .499 The percent of these females between 74 in. and 78 in. tall is %.

Solution

z = (x - mean)/ stand deviatn

For x = 74, z₁ = (74 - 66)/4 = 2

For x = 78, z₂ = (78 - 66)/4 = 3)

P( 74 < x < 78) = P(2 < z < 3) = P (z < 3) - P ( z> 2) = 0.499 - P(z< 2 ) = 0.499 - 0.477 = 0.022

% of females between 74in. and 78in. = 0.022 x 100 = 2.2%