A shipment of 11 printers contains 2 that are defective. Find the pr.pdf

•

0 j'aime•13 vues

A shipment of 11 printers contains 2 that are defective. Find the probability that a sample of 4 printers, drawn from the 11, will not contain a defective printer. Solution There are 9 nondefective printers out of the 11 total, so: (9/11)(8/10)(7/9)(6/8) = 21/55 = 0.38181818.

Formation

Contenu connexe

Plus de agrawalsales2

A set of waveforms applied to two D flip-flops is shown in Figure 4-56. These waveforms are applied to the flip-flops shown along with the values of their timing parameters. (a) List the time(s) at which there are timing violations in signal D1 for flip-flop 2. (b) List the time(s) at which there are timing violations in signal D2 for flip-flop 2. Solution flip flop 1 is a positive edge triggered flop. The inputs should be seen at the rising edge of the clock for signal D1, at the fourth rising edge i.e t = 28ns, violation occurs. D1 is rising at the same edge, it should be stable for Ts = 1ns before and Th =0.5ns after the edge. 2) flipflop 2 is a negative edge triggered. At second negative edge i.e t = 16ns and third negative edge i.e t = 24ns, D2 signal viloates the timing requirements of flop2.

A set of waveforms applied to two D flip-flops is shown in Figure 4-.pdf

agrawalsales2

A significant advantage of internal funding is the fact that retained earnings do not incur flotation cost for the company; this advantage comes at the cost of reduced capital gains for the firm’s shareholders. True False Solution A significant advantage of internal funding is the fact that retained earnings do not incur flotation cost for the company; this advantage comes at the cost of reduced capital gains for the firm’s shareholders. TRUE UNLIKE OTHER SOURCES OF FINANCEING THE RETAINED EARNNINGS DOES NOOT REQUIRE ANY ISSUE RELATED COST..

A significant advantage of internal funding is the fact that retaine.pdf

agrawalsales2

A short quiz has two true/false questions and one multiple choice question with four choices. A student guesses at each question. Assuming the choices are all equally likely, what is the probability that the student gets all three correct? Solution The events are independent, so P(A and B) = P(A)*P(B) --- P(2 correct on truefalse) = (1/2)(1/2) = 1/4 P(1 correct mult-choice) = 1/4 ---- P(this and that) = (1/4)(1/4) = 1/16.

A short quiz has two truefalse questions and one multiple choice qu.pdf

agrawalsales2

A share of common stock has just paid a dividend of $2.00 (D0 = 2.00.pdf

agrawalsales2

A set of 50 data values has a mean of 27 and a variance of 16. I. Find the standard score (z) for a data value = 19. II. Find the probability of a data value < 19. III. Find the probability of a data value > 19. Show all work. Solution I am assuming X ~ N(27, 16) to find the probability of greater than 19, just take the fact that the probabilities add to 1 and 1 - the above probability is the probability that x is greater than 19.

A set of 50 data values has a mean of 27 and a variance of 16. I. .pdf

agrawalsales2

A set of 50 data values has a mean of 21 and a variance of 16. I. Find the standard score (z) for a data value = 31. II. Find the probability of a data value > 31. III. Find the probability of a data value < 31. Show all work. Solution If the variance is 4, sigma is the sqrt(4) = 2 z = (x - mu) / sigma z = (31 - 28) / 2 z = 1.5 So from there, look at a z-table. P(z>1.5) = 1 - .9332 = .0668 P(z<1.5) = .9332.

A set of 50 data values has a mean of 21 and a variance of 16. I. .pdf

agrawalsales2

A set of 50 data values has a mean of 16 and a variance of 4. I. Find the standard score (z) for a data value = 12. II. Find the probability of a data value > 12. Solution n=50 =16 = population mean 2 = 4, so standard devation =2 A) z = (x - )/ = (12 - 16)/2 = 2 B) P(x > 12) = P( (x - )/ > (12-16)/2 ) = P( z > 2) = .0228 I did not use the standard error of the sampling distribution [ (x - )/(/n) ] because that is only used when dealing with a sample. The problems clearly state \"data value\" which I understood as it being a single observation rather than the sample mean..

A set of 50 data values has a mean of 16 and a variance of 4. I. F.pdf

agrawalsales2

A set of 10 data values has a mean and a median of $1425 and ranges .pdf

agrawalsales2

Plus de agrawalsales2 (8)