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Aliun [14]
3 years ago
9

1. Suppose half of all newborns are girls and half are boys. Hospital A, a large city hospital, records an average of 50 births

a day. Hospital B, a small, rural hospital, records an average of 10 births a day. On a particular day, which hospital is less likely to record 80% or more female births?
Mathematics
2 answers:
kicyunya [14]3 years ago
8 0

Answer:

5%

Step-by-step explanation:

Hospital A (with 50 births a day), because the more births you see, the closer the proportions will be to 0.5.

Hospital B (with 10 births a day), because with fewer births there will be less variability.

The two hospitals are equally likely to record such an event, because the probability of a boy does not depend on the number of births

Lina20 [59]3 years ago
4 0
I had the same question it’s 5%
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7 0
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7 0
3 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:

<u>Component Reliabilities:</u>

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:

<u>Reliability of the Robot with backup of component 1</u> 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

<u>Similarly, to find out the reliability of component 2:</u>

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

<u>Reliability of the Robot with backup of component 3 can be computed as:</u>

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

<u>Reliability of the Robot with backup of component 4 can be computed as:</u>

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 <u>highest reliability can be achieved by adding a backup of component 4 with a value of 0.8664</u>. So, <u>Component 4 should get the backup in order to achieve the highest reliability.</u>

<u></u>

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 <u>chances of failure</u> 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 <u>reliability for each of the component & its backup</u> 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

<u>The reliability of the robot with backups</u> 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

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

4 0
3 years ago
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