Question

1. The weights of items produced by a company are normally distributed with a mean of 5 ounces and a standard deviation of 0.2 ounces. What is the proportion of items that weigh more than 4.6 ounces?

2. What is the minimum weight of the heaviest 30.85% of all items produced?

3. Determine specifications (in ounces) that are symmetric about the mean that includes 90% of the weight of all items produced.

Answer 1

Solution :

Given that ,

mean = = 5

standard deviation = = 0.2

1) P(x > 4.6) = 1 - p( x< 4.6)

=1- p P[(x - ) / < (4.6 - 5) / 0.2]

=1- P(z < -2.0)

Using z table,

= 1 - 0.0228

= 0.9772

2) Using standard normal table,

P(Z > z) = 30.85%

= 1 - P(Z < z) = 0.3085  

= P(Z < z) = 1 - 0.3085

= P(Z < z ) = 0.0.6915

= P(Z < 0.50 ) = 0.6915  

z = 0.50

Using z-score formula,

x = z * +

x = 0.50 * 0.2 + 5

x = 5.1 ounces

3) Using standard normal table,

P( -z < Z < z) = 90%

= P(Z < z) - P(Z <-z ) = 0.90

= 2P(Z < z) - 1 = 0.90

= 2P(Z < z) = 1 + 0.90

= P(Z < z) = 1.90 / 2

= P(Z < z) = 0.95

= P(Z < 1.645) = 0.95

= z  ± 1.645

Using z-score formula,

x = z * +

x = -1.645 * 0.2 + 5

x = 4.67 ounces

Using z-score formula,

x = z * +

x = 1.645 * 0.2 + 5

x = 5.33 ounces

90% symmetric values is 4.67 ounces and 5.33 ounces.