The lengths of a certain component are normally distributed with a mean of 250 m

Question

The lengths of a certain component are normally distributed with a mean of 250 mm and a standard deviation of 0.6 mm. Four of the components are joined end-to-end and fitted into a slot.

1) If the slots have length 1002.5 mm, what proportion of the time will four randomly selected components be able to fit into the slot?

2) Suppose that the lengths of the slots are normally distributed with a mean of 1002.5 mm and a standard deviation of 0.4 mm. What proportion of the time will four randomly selected components be able to fit into the slot?

Explanation / Answer

Lenght of a certain componenet are normally distributed with mean = 250 mm and standered deivation = 0.6 mm

four of the components are fitted into a slot

(1) slots have length = 1002.5 mm

Mean length of the 4 componets = 4 * 250 = 1000 mm

standard deviation for 4 slots = sqrt ( 4 * 0.62) = 1.2 mm

So slots have length = 1002.5 mm

Z- value = (1002.5 - 1000)/1.2 = 2.0833

so P ( X< 1002.5; 1000; 0.6) = 0.9813 ( from Z - table)

so 98.13 % time, randonmly selected components be able to fit into the slot.

(2) Here length of slots are 1002.5 mm and with standerd deviation = 0.4 mm

so pooled standerd deviation betwen 4 components and the slot = sqrt ( 1.22 + 0.42) = 1.265

so Z - value = ( 1002.5 - 1000) / 1.265 = 1.9762

so Calculating P - value from Z - value P ( X< 1002.5 ; 1000; 1.2) = 0.9759

so it will be 97.59 % time that four randomly selected components be able to fit into the slot.

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