Question

The quality control department at Alexander Fasteners measured the length of 100 bolts randomly selected from a specified order. The mean length was found to be 6.65 cm, and the standard deviation was 0.5 cm. The bolt lengths are normally distributed.

(a) Determine the percentage of bolts shorter than 5.79 cm.

(b) Determine the percentage of bolts longer than 7.46 cm.

(c) Determine the percentage of bolts shorter than 5.79 cm.

(d) Determine the percentage of bolts that are longer than the nominal length of 6.65 cm.

Please show all work and provide diagrams.

Answer 1

Solution :

Given that ,

mean = = 6.65

standard deviation = = 0.5

(a)

P(x < 5.79) = P[(x - ) / < (5.79 - 6.65) / 0.5]

= P(z < -1.72)

= 0.0427

percentage = 4.27%

(b)

P(x > 7.46) = 1 - P(x < 7.46)

= 1 - P[(x - ) / < (7.46 - 6.65) / 0.5)

= 1 - P(z < 1.62)

= 1 - 0.9474

= 0.0526

percentage = 5.26%

(c)

P(x < 5.79) = P[(x - ) / < (5.79 - 6.65) / 0.5]

= P(z < -1.72)

= 0.0427

percentage = 4.27%

(d)

P(x > 6.65) = 1 - P(x < 6.65)

= 1 - P[(x - ) / < (6.65 - 6.65) / 0.5)

= 1 - P(z < 0)

= 1 - 0.5

= 0.5

percentage = 0.05%