Thanks for looking.

I appreciate your time and effort.

Ok so N = 8Suppose that 65% of the families in a town own computers. If eight families are surveyed at random, what is the prob. that at least four own computers ?

X = 4

Success = 0.65 ?

Failure = 0.35 ?

I am using the formulate :

$\displaystyle P(X) = 8C4 * 0.65 ^4 * 0.35^0$

I know i am doing something wrong...

The wording is really messing me up....Trump - airlines has dtermined 5% of its customers do not show up for their flights.

If a passenger is bumped off a flight because of overbooking, the airline pays the customer $200. What is the expected payout by the airline if it overbooks a 240 seat airplane by 5%

0.3In a binomial dist., if 10 trials are conducted, and the prb. of failure is 0.3, what is the expected value ?

3

0.7

7

None

N = 10

Q = 0.3

P = 0.7

$\displaystyle E(X) = n * p$

$\displaystyle E(X) = 10 * 0.7$

$\displaystyle E(X) = 7$

Is that correct ?

Ok thanks for your time so far...

This is the hardest...

I am twisted in there...A manufacturer of electronics components produces precision resistors designd to have a tolerance of +/- 1%. from quality control testing, the manu. knows that about 1 resistor in 6 is actually within just 0.3% of its nominal value. A customer needs 5 of these more precise resistors. what is the prob. of finding exactly 5 such resistors aming the first 8 tested?

There are 2 of them....

1 in 6 and 5 in 8...