The time required to assemble an electronic component is normally distributed, with a mean of 12 minutes and a standard deviation of 1.5 minutes. Find the probability that a particular assembly take less than 10 minutes.
a. |
0.6542 |
|
b. |
0.0918 |
|
c. |
0.8164 |
|
d. |
0.9082 |
|
e. |
0.4541 |
Solution :
Given that,
mean = = 12
standard deviation = = 1.5
P(X< 10) = P[(X- ) / < (10-12) /1.5 ]
= P(z <-1.33 )
Using z table
=0.0918
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