Here are the historical probabilities p(x) = P(X = x) that a fixed length of cast-iron piping will be manufactured and be found, on inspection, to have various numbers X of potentially weakening defects:x 0 1 2 3 4p(x) .45 .21 .13 .12 ,09F(x)Calculate the expected number of defects in a pipe E(X).

Step-by-step explanation:

For this case we have the probability density function given:

X          0        1         2       3        4

P(X)  0.45   0.21    0.13   0.12  0.09

The cdf of F(X) would be given by:

X          0         1         2        3     4

F(X)   0.45   0.66   0.79   0.91   1

The expected value of a random variable X is the n-th moment about zero of a probability density function f(x) if X is continuous, or the weighted average for a discrete probability distribution, if X is discrete.

And we can use the following formula in order to calculate the expected value:

And replacing we got:

step-by-step explanation: yes

the largest number of the four options is 1.8 x 10-2

you have it in scientific notation.

1.80 x 10-6   = 0.00000180

1.80 x 10-4   = 0.000180 (bigger, or "less small", than the former one)

so

1.8 x 10-2       = 0.0180. as you see. the "less small", or the biggest one of them all.

as you have negative exponent, it means you are dividing 1.8 by 100 (i.e. moving the dot towards left, two times).

if you had positive exponent for the scientific notation, power of tenths, you would be multiplying times 100 (moving the dot towards right) as below:

1.8 x (10)2 = 180

so. if you have negative exponent in scientific notation, move the dot toward left. if it is positive, toward right as many times the exponent tells you.