Question

2. Women’s hands are 6.9 inches on average (from the tip of the middle finger to the heel of the hand). The standard deviation is 0.34 inches. Assume that hand length is normally distributed. A. What is the probability that a woman would have a hand length longer than 7.58 inches? Show your work. Start by finding the z-score for 7.58 and then use what you know about the normal distribution to find the probability. B. What is the probability that a woman would have a hand length shorter than 6.56 inches? Show your work. C. What is the probability that a woman would have a hand length between 6.22 and 7.58 inches? D. If you were a glove maker, how many pairs out of a thousand would you make so that the hand length would be between 6.22 and 7.58 inches?

Answer 1

Part a)

P ( X > 7.58 ) = 1 - P ( X < 7.58 )
Standardizing the value

Z = ( 7.58 - 6.9 ) / 0.34
Z = 2

P ( Z > 2 )
P ( X > 7.58 ) = 1 - P ( Z < 2 )
P ( X > 7.58 ) = 1 - 0.9772
P ( X > 7.58 ) = 0.0228

Part b)

P ( X < 6.56 )
Standardizing the value

Z = ( 6.56 - 6.9 ) / 0.34
Z = -1

P ( X < 6.56 ) = P ( Z < -1 )
P ( X < 6.56 ) = 0.1587

Part c)

P ( 6.22 < X < 7.58 )
Standardizing the value

Z = ( 6.22 - 6.9 ) / 0.34
Z = -2
Z = ( 7.58 - 6.9 ) / 0.34
Z = 2
P ( -2 < Z < 2 )
P ( 6.22 < X < 7.58 ) = P ( Z < 2 ) - P ( Z < -2 )
P ( 6.22 < X < 7.58 ) = 0.9772 - 0.0228
P ( 6.22 < X < 7.58 ) = 0.9545

Part d)

N = 1000, P = 0.9545

P ( 6.22 < X < 7.58 ) = 0.9545

n = 1000 * 0.9545 = 954.5   955