The breaking strength of a cable is assumed to be lognormally distributed with a mean of 75 and standard deviation of 20. • if the load of magnitude 50 is hung from the cable, determine the probability of failure of the cable? • if the load can take values 30, 50, and 70 with probabilities 0.2, 0.35, and 0.45 respectively, determine the probability of failure of the cable?

If the load of magnitude 50 is hung from the cable, determine the probability of failure of the cable?

There is a 10.56% probability of failure of the cable.

If the load can take values 30, 50, and 70 with probabilities 0.2, 0.35, and 0.45 respectively, determine the probability of failure of the cable?

There is a 15.87% probability of failure of the cable.

Step-by-step explanation:

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean and standard deviation , the zscore of a measure X is given by:

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

The breaking strength of a cable is assumed to be lognormally distributed with a mean of 75 and standard deviation of 20. This means that .

If the load of magnitude 50 is hung from the cable, determine the probability of failure of the cable?

This is the pvalue of Z when .

has a pvalue of 0.1056.

There is a 10.56% probability of failure of the cable.

If the load can take values 30, 50, and 70 with probabilities 0.2, 0.35, and 0.45 respectively, determine the probability of failure of the cable?

Now we have that

has a pvalue of 0.1587.

There is a 15.87% probability of failure of the cable.

ok

step-by-step explanation:

geometric series converge the common ratio r has magnitude less than one.

which is equivalent to

for the example we need

|2-x| < 1

-1 < 2-x   and 2-x < 1

x < 3 and x > 1

we can write that

answer: 1 < x < 3

1a. common ratio r=250/25=10 > 1     diverges

1b.   r=50/100=1/2 converges

when these converge they converge to

where a is the first term.

answer: 200

1c. r=-1   so it's not the case that |r| < 1.     diverges

2a. x is the common ratio so we converge when |x|< 1

2b.   r=-10x / -2 = 5x.

we converge when

|5x| < 1

|x| < 1/5

2c.   r=3-2x

-1 < 3-2x and 3-2x < 1

-4 < -2x and 2 < 2x

x < 2 and 1 < x

1 < x < 2

2d. [/tex] r=-48x^3/12 = 4x^3[/tex]

|4x^3| < 1

|x^3| < \frac 1 4

|x| < 4^{-1/3}