Question

Please give the detailed solution

The diameter of Golf balls manufactured at a large factory in Trenton N.J. is expected to be approximately normally distributed with a mean of 1.30 inches and a standard deviation of 0.04 inches. What is the probability that a randomly selected Golf ball will have a diameter of: a. Between 1.28 and 1.30 inches? b. Between 1.31 and 1.33 inches? c. Between what two values, in terms of the diameter, will 60% of the Golf balls fall? If random samples of 16 Golf balls were selected, d. What proportion of the sample means would be between 1.28 and 1.30 inches? e. Compare your results in (a) and (d) .

Answer 1

a) P(1.28 < X < 1.30)

= P((1.28 - )/ < (X - )/ < (1.30 - )/)

= P((1.28 - 1.30)/0.04 < Z < (1.30 - 1.30)/0.04)

= P(-0.5 < Z < 0)

= P(Z < 0) - P(Z < -0.5)

= 0.5 - 0.3085

= 0.1915

b) P(1.31 < X < 1.33)

= P((1.31 - )/ < (X - )/ < (1.33 - )/)

= P((1.31 - 1.30)/0.04 < Z < (1.33 - 1.30)/0.04)

= P(0.25 < Z < 0.75)

= P(Z < 0.75) - P(Z < 0.25)

= 0.7734 - 0.4013

= 0.3721

c) P(-x < X < x) = 0.60

or, P(-z < Z < z) = 0.60

or, P(Z < z) - P(Z < -z) = 0.60

or, P(Z < z) - (1 - P(Z < z)) = 0.60

or, P(Z < z) - 1 + P(Z < z) = 0.60

or, 2P(Z < z) = 1.60

or, P(Z < z) = 0.8

or, z = 0.84

or, (x - )/ = 0.84

or, (x - 1.30)/0.04 = 0.84

or, x = 0.84 * 0.04 + 1.30

or, x = 1.33

d) P(1.28 < < 1.30)

= P((1.28 - )/() < ( - )/() < (1.30 - )/())

= P((1.28 - 1.30)/(0.04/sqrt(16)) < Z < (1.30 - 1.30)/(0.04/sqrt(16)))

= P(-2 < Z < 0)

= P(Z < 0) - P(Z < -2)

= 0.5 - 0.0228

= 0.4772

e) As the sample size in part (d) increases, the probability in part (d) is greater than the probability in part (a).