Answer to Question #295269 in Statistics and Probability for Hon

Question

Suppose the force acting on a column that helps to support a building is normally

distributed with mean 16.0 kips and standard deviation 1.35 kips. What is the probability that the

force

a. Is at most 17 kips?

b. Is between 11 and 13 kips?

c. Differs from 16.0 kips by at most 1.6 standard deviations?

Answer 1

$\mu=16\\\sigma=1.35$

$a)$

We determine the probability,

$p(x\le17)=p({x-\mu\over \sigma}\lt{17-\mu\over\sigma})=p(Z\le{17-16\over1.35})=p(Z\le0.74)=\phi(0.74)=0.7704$

$b)$

We find,

$p(11\lt x\lt 13)=p({11-16\over1.35}\lt Z\lt{13-16\over1.35})=p(-3.70\lt Z\lt -2.22)=\phi(-2.22)-\phi(-3.70)=0.0132-0.00011=0.01309$

$c)$

We need to determine the upper and lower limit which are 1.6 standard deviations from the 16.

Lower limit is,

$16-(1.6\times1.35)=16-2.16=13.84$

Upper limit is,

$16+(1.6\times1.35)=16+2.16=18.16$

Now we have to find the probability,

$p(13.84\lt x\lt18.16)=p({13.84-16\over1.35}\lt Z\lt{18.16-16\over1.35})=p(-1.6 \lt Z\lt 1.6)=\phi(1.6)-\phi(-1.6)=0.9452-0.0548=0.8904$

Therefore, the probability that the force differs from 16.0 kips by at most 1.6 standard deviations is 0.8904.