Problem 1
A physiotherapist with a male football team is interested in studying the relationship between foot injuries and the positions at which the players play from the data collected
Striker Forward Attacking Midfielder Winger Total
Players Injured 45 56 24 20 145
Players Not Injured 32 38 11 9 90
Total 77 94 35 29 235
1.1 What is the probability that a randomly chosen player would suffer an injury?
-The total number of players are 235.
-The total number of players Injured are 145
-The probability that a randomly chosen player would suffer an injury is 145/235= 0.62 or 62%
1.2 What is the probability that a player is a forward or a winger?
-The total number of players are 235.
-The total number of players playing as a forward are 94
-The total number of players playing as a winger are 29
-The probability that a player is a forward or a winger 0.52 or 52.34%
1.3 What is the probability that a randomly chosen player plays in a striker position and has a foot injury?
-The total number of players which are strikers are 77
-The total number players which are striker and have injury are 45
So, the probability for player which are strikers and have foot injury are 45/77 which is 0.5844 or 58.44%
The probability that a randomly chosen player plays in a striker position and has a foot injury is 0.19
1.4 What is the probability that a randomly chosen injured player is a striker?
-The Total number of injured players are 145
-Total number of players which are injured and striker are 45
So, the probability of player chosen which are injured and is a striker is 45/145 which is 0.31 or 31%
1.5 What is the probability that a randomly chosen injured player is either a forward or an attacking midfielder?
-The total number of injured players are 145
-Total number of players which are injured and forward and midfielder are 56 and 24 respectively.
So, the probability of players chosen injured and are forward and midfielder are (56+24)/145 which is 0.5517 or 55.17%
Problem 2
An independent research organization is trying to estimate the probability that an accident at a nuclear power plant will result in radiation leakage. The types of accidents possible at the plant are, fire hazards, mechanical failure, or human error. The research organization also knows that two or more types of accidents cannot occur simultaneously.
According to the studies carried out by the organization, the probability of a radiation leak in case of a fire is 20%, the probability of a radiation leak in case of a mechanical 50%, and the probability of a radiation leak in case of a human error is 10%. The studies also showed the following;
• The probability of a radiation leak occurring simultaneously with a fire is 0.1%.
• The probability of a radiation leak occurring simultaneously with a mechanical failure is 0.15%.
• The probability of a radiation leak occurring simultaneously with a human error is 0.12%.
On the basis of the information available, answer the questions below:
GIVEN-
The probability of a radiation leak in case of a fire is 20%,
The probability of a radiation leak in case of a mechanical 50%,
The probability of a radiation leak in case of a human error is 10%
The probability of a radiation leak occurring simultaneously with a fire is 0.1%
The probability of a radiation leak occurring simultaneously with a mechanical failure is 0.15%
The probability of a radiation leak occurring simultaneously with a human error is 0.12
2.1 What are the probabilities of a fire, a mechanical failure, and a human error respectively?
Probability of Fire
= Probability of a radiation leak in case of a fire/probability of a radiation leak in case of a fire
= 0.10/20
=0.005
Probability of a mechanical failure
=Probability of a radiation leak in case of a Mechanical failure / probability of a radiation leak in case of a Mechanical failure
= 0.15/50
=0.003
Probability of Human Error
probability of a radiation leak in case of a Human Error / probability of a radiation leak in case of a Human Error
= 0.15/50
=0.012
2.2 What is the probability of a radiation leak?
Probability of a radiation leak = The probability of a radiation leak occurring simultaneously with a fire + probability of a radiation leak occurring simultaneously with a mechanical failure + The probability of a radiation leak occurring simultaneously with a human error
Probability of a radiation leak = 0.1+0.15+0.12 = 0.37
Probability of Radiation Leak is 0.37
2.3 Suppose there has been a radiation leak in the reactor for which the definite cause is not known. What is the probability that it has been caused by:
• A Fire.
• A Mechanical Failure.
• A Human Error.
A-Probability that it has been caused by Fire = probability of a radiation leak in case of a fire/Probability of a radiation leak
Probability that it has been causedby Fire = 0.1/0.37
Probability that it has been caused by Fire is 0.27
B- Probability that it has been caused by Mechanical failure = probability of a radiation leak in case of a mechanical failure/Probability of a radiation leak
Probability that it has been caused by Mechanical failure = 0.15/0.37
Probability that it has been caused by Mechanical failure is 0.41
Problem 3:
The breaking strength of gunny bags used for packaging cement is normally distributed with a mean of 5 kg per sq. centimeter and a standard deviation of 1.5 kg per sq. centimeter. The quality team of the cement company wants to know the following about the packaging material to better understand wastage or pilferage within the supply chain; Answer the questions below based on the given information; (Provide an appropriate visual representation of your answers, without which marks will be deducted)
Given
• Gunny bags used for packaging cement is normally distributed
• Mean is 5 kg per sq. centimetre
• Standard deviation is 1.5 kg per sq. centimetre
3.1 What proportion of the gunny bags have a breaking strength less than 3.17 kg per sq cm?
11.12% proportion of the gunny bags have a breaking strength less than 3.17 kg per sq cm
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