s) The lifetimes of a certain type of light bulbs (in years) have pt distributio
ID: 2925032 • Letter: S
Question
Explanation / Answer
13. We have been given params of the normal distribution :
Mean = 202
Stdev = 3
a. P(X>205) = P(Z> 205-202/3) = P(Z>1) = .1587
b. P(x<c) = .15, so, c-202 / 3 = -1.035, So, c = -1.035*3+202 = 198.895
c. New mean = 200g, stdev = 3 gm. Only 1% are underweight if
Z = 200-205/3 = -5/3, P(Z=-5/3) = P(Z=-1.67) = .0475. 4.75% are underweight. Yes, you need to reset the filling amount so that only 1% of the packets are underweight.
To make only 1% underweight we have a Z of -2.33 ,which is X-205/3 = -2.33
X = -2.33*3+205 = 198g, should be the stamped weight on the pocket so that only 1% is underweight
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