PLEASE HELP ME 30 POINTS

A coin will be flipped repeatedly until the sequence TTH (tail/tail/head) comes up. Successive flips are independent, and the co

padilas [110]

Answer:

$P = \frac{1}{3}$

Step-by-step explanation:

The calculation of the value of p minimizes is shown below:-

We will assume the probability of coming heads be p

p(H) = p

p(T) = 1 - P

Now, H and T are only outcomes of flipping a coin

So,

P(TTH) = (1 - P) = (1 - P) (1 - P) P

= (1 + P^2 - 2 P) P

= P^3 - 2P^2 + P

In order to less N,TTH

we need to increase P(TTH)

The equation will be

$\frac{d P(TTH)}{dP} = 0$

3P^2 - 4P + 1 = 0

(3P - 1) (P - 1) = 0

P = 1 and 1 ÷ 3

For P(TTH) to be maximum

$\frac{d^2 P(TTH)}{dP} < 0 for\ P\ critical\\\\\frac{d (3P^2 - 4P - 1)}{dP}$

= 6P - 4

and

(6P - 4) is negative which is for

$P = \frac{1}{3}$

5 0

2 years ago

One of the industrial robots designed by a leading producer of servomechanisms has four major components. Components’ reliabilit

Ivahew [28]

Answer:

a) Reliability of the Robot = 0.7876

b1) Component 1: 0.8034

Component 2: 0.8270

Component 3: 0.8349

Component 4: 0.8664

b2) Component 4 should get the backup in order to achieve the highest reliability.

c) Component 4 should get the backup with a reliability of 0.92, to obtain the highest overall reliability i.e. 0.8681.

Step-by-step explanation:

Component Reliabilities:

Component 1 (R1) : 0.98

Component 2 (R2) : 0.95

Component 3 (R3) : 0.94

Component 4 (R4) : 0.90

a) Reliability of the robot can be calculated by considering the reliabilities of all the components which are used to design the robot.

Reliability of the Robot = R1 x R2 x R3 x R4

= 0.98 x 0.95 x 0.94 x 0.90

Reliability of the Robot = 0.787626 ≅ 0.7876

b1) Since only one backup can be added at a time and the reliability of that backup component is the same as the original one, we will consider the backups of each of the components one by one:

Reliability of the Robot with backup of component 1 can be computed by first finding out the chance of failure of the component along with its backup:

Chance of failure = 1 - reliability of component 1

= 1 - 0.98

= 0.02

Chance of failure of component 1 along with its backup = 0.02 x 0.02 = 0.0004

So, the reliability of component 1 and its backup (R1B) = 1 - 0.0004 = 0.9996

Reliability of the Robot = R1B x R2 x R3 x R4

= 0.9996 x 0.95 x 0.94 x 0.90

Reliability of the Robot = 0.8034

Similarly, to find out the reliability of component 2:

Chance of failure of component 2 = 1 - 0.95 = 0.05

Chance of failure of component 2 and its backup = 0.05 x 0.05 = 0.0025

Reliability of component 2 and its backup (R2B) = 1 - 0.0025 = 0.9975

Reliability of the Robot = R1 x R2B x R3 x R4

= 0.98 x 0.9975 x 0.94 x 0.90

Reliability of the Robot = 0.8270

Reliability of the Robot with backup of component 3 can be computed as:

Chance of failure of component 3 = 1 - 0.94 = 0.06

Chance of failure of component 3 and its backup = 0.06 x 0.06 = 0.0036

Reliability of component 3 and its backup (R3B) = 1 - 0.0036 = 0.9964

Reliability of the Robot = R1 x R2 x R3B x R4

= 0.98 x 0.95 x 0.9964 x 0.90

Reliability of the Robot = 0.8349

Reliability of the Robot with backup of component 4 can be computed as:

Chance of failure of component 4 = 1 - 0.90 = 0.10

Chance of failure of component 4 and its backup = 0.10 x 0.10 = 0.01

Reliability of component 4 and its backup (R4B) = 1 - 0.01 = 0.99

Reliability of the Robot = R1 x R2 x R3 x R4B

= 0.98 x 0.95 x 0.94 x 0.99

Reliability of the Robot = 0.8664

b2) According to the calculated values, the highest reliability can be achieved by adding a backup of component 4 with a value of 0.8664. So, Component 4 should get the backup in order to achieve the highest reliability.

c) 0.92 reliability means the chance of failure = 1 - 0.92 = 0.08

We know the chances of failure of each of the individual components. The chances of failure of the components along with the backup can be computed as:

Component 1 = 0.02 x 0.08 = 0.0016

Component 2 = 0.05 x 0.08 = 0.0040

Component 3 = 0.06 x 0.08 = 0.0048

Component 4 = 0.10 x 0.08 = 0.0080

So, the reliability for each of the component & its backup is:

Component 1 (R1BB) = 1 - 0.0016 = 0.9984

Component 2 (R2BB) = 1 - 0.0040 = 0.9960

Component 3 (R3BB) = 1 - 0.0048 = 0.9952

Component 4 (R4BB) = 1 - 0.0080 = 0.9920

The reliability of the robot with backups for each of the components can be computed as:

Reliability with Component 1 Backup = R1BB x R2 x R3 x R4

= 0.9984 x 0.95 x 0.94 x 0.90

Reliability with Component 1 Backup = 0.8024

Reliability with Component 2 Backup = R1 x R2BB x R3 x R4

= 0.98 x 0.9960 x 0.94 x 0.90

Reliability with Component 2 Backup = 0.8258

Reliability with Component 3 Backup = R1 x R2 x R3BB x R4

= 0.98 x 0.95 x 0.9952 x 0.90

Reliability with Component 3 Backup = 0.8339

Reliability with Component 4 Backup = R1 x R2 x R3 x R4BB

= 0.98 x 0.95 x 0.94 x 0.9920

Reliability with Component 4 Backup = 0.8681

Component 4 should get the backup with a reliability of 0.92, to obtain the highest overall reliability i.e. 0.8681.

4 0

2 years ago

The number of surface flaws in plastic panels used in the interior of automobiles has a Poisson distribution with a mean of 0.06

True [87]

Answer:

a) 0.7403

b)0.0498

c)0.2240

Step-by-step explanation:

Given: The number of surface flaws in plastic panels used in the interior of automobiles has a Poisson distribution

We know that to calculate expected number of flaws use

expected number of flaws =10×0.03 =0.3

a) probability that there are no surface flaws in an auto's interior =P(X=0)=

e^-0.3 =0.7408

b) probability that none of the 10 cars has any surface flaws =(e^-0.3)^10 =0.0498

c) probability that at most one car has any surface flaws =P(X<=1)=P(X=0)+P(X=1)

this means

=10C_0(1-0.7408)^0(0.7408)^10+10C_1(1-0.7408)^1(0.7408)^9=0.2240

4 0

1 year ago

Read 2 more answers

Which linear equation has no solution? 4x+7=3x+7

ludmilkaskok [199]

Answer:

x=0

Step-by-step explanation:

4x+7=3x+7

4x-3x=7-7

x=0

7 0

1 year ago

Is rap music more popular among young Indians than among young Asians? A sample survey compared 634 randomly chosen Indians aged

BabaBlast [244]

Answer:

Step-by-step explanation:

Confidence interval for the difference in the two proportions is written as

Difference in sample proportions ± margin of error

Sample proportion, p = x/n

Where x = number of successes

n = number of samples

For the indians,

x = 368

n1 = 634

p1 = 368/634 = 0.58

For the asians

x = 130

n2 = 567

p2 = 130/567 = 0.23

Margin of error = z√[p1(1 - p1)/n1 + p2(1 - p2)/n2]

To determine the z score, we subtract the confidence level from 100% to get α

α = 1 - 0.95 = 0.05

α/2 = 0.05/2 = 0.025

This is the area in each tail. Since we want the area in the middle, it becomes

1 - 0.025 = 0.975

The z score corresponding to the area on the z table is 1.96. Thus, confidence level of 95% is 1.96

Margin of error = 1.96 × √[0.58(1 - 0.58)/634 + 0.23(1 - 0.23)/567]

= 1.96 × √0.00038422713 + 0.00031234568)

= 0.052

Confidence interval = 0.58 - 0.23 ± 0.052

Confidence interval = 0.35 ± 0.052

8 0

1 year ago