Specifications for a part for a DVD player state that the pa

Specifications for a part for a DVD player state that the part should weigh between 24.3 and 25.3 ounces. The process that produces the parts has a mean of 24.8 ounces and a standard deviation of .23 ounce. The distribution of output is normal.

a. What percentage of parts will not meet the weight specs? (Round your \"z\" value and final answer to 2 decimal places. Omit the \"%\" sign in your response.) Percentage of parts %

b. Within what values will 99.74 percent of sample means of this process fall, if samples of n = 6 are taken and the process is in control (random)?

Solution

It is a problem of Normal Distribution.

As per the formula, Z Score = (x - M)/ Standard deviation

   here, X = Actual value

M = Mean Value

As pe the given value, X1 =24.7 X2 =25.7 and M = 25.2

so Z1 = (x1-M)/.2 = (24.7 - 25.2)/.2 = -2.5

Z2= (x2-M)/.2 = ( 25.7 - 25.2)/.2 = +2.5

As per the given table of Z scores, 2.5 means area of .4938

since Z value ranges from -2.5 to 2.5 it means that it will cover total area of 2*.4938 = .9876

It means that Products who will not meet specifications = 1-.9876 = .012

so % of products not meetings specifications =1.20

2.

To get 99.74% result

Area in each side = .9974/2 = .4987

to get the area of eachsside as .4987 ( from the given table),  Z value should range from -3 to + 3

For z1= -3 = (x1-M)/.2 with M=25.2

after solving above, X1 = 24.6

For z2= +3 = (x2-M)/.2 with M=25.2

after solving above, X2 = 25.8

so lower value=24.6

Higher value= 25.8