Question

3) The breaking strengths of nylon fibers in dynes are normally distributed with a mean of 11951 and a variance of 200118. What is the probability that a fiber strength is more than 12450? Round your answers to the nearest thousandth.

4) A new 1.5-volt battery has an actual voltage that is uniformly distributed between 1.43 and 1.79 volts. Estimate the probability that the sum of the voltages from 119 new batteries lies between 179 and 189 volts. Round your answers to the nearest thousandth.

Answer 1

3)ANSWER:-

$mean=11951$

$variance=200118$

$standard deviation=447.3455$

Here, we have to calculate the probability of fiber  more than 12450:-

$P(X>12450)=1-P(z<12450)$

$P(X>12450)=1-P(z<\frac{x-\mu }{s})$

$P(X>12450)=1-P(z<\frac{12450-11951}{447.3455})$

$P(X>12450)=1-P(z<\frac{499}{447.3455})$

$P(X>12450)=1-P(z<1.12)$

$P(X>12450)=1-0.8686$

$P(X>12450)=0.1314$

$P(X>12450)=0.1$$nearly$$thousandth$$value$.
NOTE:-

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