Question

Assume that the widths of the metal fasteners manufactured by a company are normally distributed with a mean of 6.00 cm and a standard deviation of .05cm.

a) What percentage of the metal fasteners have a width less than 5.88 cm?

b) what percentage of the metal fasteners have widths between 5.94 and 6.06cm?

c) For what width are only 7% of all metal fasteners that wide or wider?

I know the answers, but I need the explanation because I don't know how to do it. Thank you.

Answer 1

Given,

= 6.00, = 0.05

We convert this to standard normal as

P(X < x) = P( Z < x - / )

a)

P( X < 5.88) = P( Z < 5.88 - 6 / 0.05)

= P( Z < -2.4)

= 0.0082

b)

P(5.94 < X < 6.06) = P( X < 6.06) - P( X < 5.94)

= P( Z < 6.06 - 6 / 0.05) - P( Z < 5.94 - 6.00 / 0.05)

= P( Z < 1.2) - P( Z < -1.2)

= 0.8849 - 0.1150

= 0.7699

c)

We have to calculate x such that

P( X > x) = 0.07

P( X < x) = 0.93

P( Z < x - / ) = 0.93

From the Z table, z-score for the probability of 0.93 is 1.4758

x - / = 1.4758

x - 6 / 0.05 = 1.4758

Solve for x

x = 6.0738 cm