Question 1: 150
Question 2: 50
Question 3: 58% chance
Question 4: Choosing a set of parallel side on a polygon.
Question 5: 1/6
Question 6: 1/5
Question 7: ?
Question 8: Rolling a number greater than 1
Question 9: 1/12
Question 10: 30
Question 11: ?
Question 12: A green marble is selected.
Hope I helped I tried as much as possible.
Question 1: 150
Question 2: 50
Question 3: 58% chance
Question 4: Choosing a set of parallel side on a polygon.
Question 5: 1/6
Question 6: 1/5
Question 7: ?
Question 8: Rolling a number greater than 1
Question 9: 1/12
Question 10: 30
Question 11: ?
Question 12: A green marble is selected.
Hope I helped I tried as much as possible.
See explanation
Step-by-step explanation:
Q1)a
- Denote a random variable ( X ) as the life time of a brand of bulb produced.
- The given mean ( μ ) = 210 hrs and standard deviation ( σ ) = 56 hrs. The distribution is symbolized as follows:
X ~ Norm ( 210 , 56^2 )
i) The bulb picked to have a life time of at least 300 hours.
- We will first standardize the limiting value of the RV ( X ) and determine the corresponding Z-score value:
P ( X ≥ x ) = P ( Z ≥ ( x - μ ) / σ )
P ( X ≥ 300 ) = P ( Z ≥ ( 300 - 210 ) / 56 )
P ( X ≥ 300 ) = P ( Z ≥ 1.607 )
- Use the standard normal look-up table for limiting value of Z-score:
P ( X ≥ 300 ) = P ( Z ≥ 1.607 ) = 0.054 .. Answer
ii) The bulb picked to have a life time of at most 100 hours.
- We will first standardize the limiting value of the RV ( X ) and determine the corresponding Z-score value:
P ( X ≤ x ) = P ( Z ≤ ( x - μ ) / σ )
P ( X ≤ 100 ) = P ( Z ≤ ( 100 - 210 ) / 56 )
P ( X ≤ 100 ) = P ( Z ≤ -1.9643 )
- Use the standard normal look-up table for limiting value of Z-score:
P ( X ≤ 100 ) = P ( Z ≤ -1.9643 ) = 0.0247 .. Answer
iii) The bulb picked to have a life time of between 150 and 250 hours.
- We will first standardize the limiting value of the RV ( X ) and determine the corresponding Z-score value:
P ( x1 ≤ X ≤ x2 ) = P ( ( x1 - μ ) / σ ≤ Z ≤ ( x2 - μ ) / σ )
P ( 150 ≤ X ≤ 250 ) = P ( ( 150 - 210 ) / 56 ≤ Z ≤ ( 250 - 210 ) / 56 )
P ( 150 ≤ X ≤ 250 ) = P ( -1.0714 ≤ Z ≤ 0.71428 )
- Use the standard normal look-up table for limiting value of Z-score:
P ( 150 ≤ X ≤ 250 ) = P ( -1.0714 ≤ Z ≤ 0.71428 ) = 0.6205 .. Answer
Q1)b
- Denote event (A) : Kofi solves the problem correctly. Then the probability of him answering successfully is:
p ( A ) = 0.25
- Denote event (B) : Menesh solves the problem correctly. Then the probability of him answering successfully is:
p ( B ) = 0.4
- The probability that neither of them answer the question correctly is defined by a combination of both events ( A & B ). The two events are independent.
- So for independent events the required probability can be stated as:
p ( A' & B' ) = p ( A' ) * p ( B' )
p ( A' & B' ) = [ 1 - p ( A ) ] * [ 1 - p ( B ) ]
p ( A' & B' ) = [ 1 - 0.25 ] * [ 1 - 0.4 ]
p ( A' & B' ) = 0.45 ... Answer
Q2)a
- A discrete random variable X: defines the probability of getting each number on a biased die.
- From the law of total occurrences. The sum of probability of all possible outcomes is always equal to 1.
∑ p ( X = xi ) = 1
p ( X = 1 ) + p ( X = 2 ) + p ( X = 3 ) + p ( X = 4 ) + p ( X = 5 ) + p ( X = 6 )
1/6 + 1/6 + 1/5 + k + 1/5 + 1/6 = 1
k = 0.1 ... Answer
- The expected value E ( X ) or mean value for the discrete distribution is determined from the following formula:
E ( X ) = ∑ p ( X = xi ) . xi
E ( X ) = (1/6)*1 + (1/6)*2 + (1/5)*3 + (0.1)*4 + (1/5)*5 + (1/6)*6
E ( X ) = 3.5 .. Answer
- The expected-square value E ( X^2 ) or squared-mean value for the discrete distribution is determined from the following formula:
E ( X^2 ) = ∑ p ( X = xi ) . xi^2
E ( X^2 ) = (1/6)*1 + (1/6)*4 + (1/5)*9 + (0.1)*16 + (1/5)*25 + (1/6)*36
E ( X^2 ) = 15.233 .. Answer
- The variance of the discrete random distribution for the variable X can be determined from:
Var ( X ) = E ( X^2 ) - [ E ( X ) ] ^2
Var ( X ) = 15.2333 - [ 3.5 ] ^2
Var ( X ) = 2.9833 ... Answer
- The cumulative probability of getting any number between 1 and 5 can be determined from the sum:
P ( 1 < X < 5 ) = P ( X = 2 ) + P ( X = 3 ) + P ( X = 4 )
P ( 1 < X < 5 ) = 1/6 + 1/5 + 0.1
P ( 1 < X < 5 ) = 0.467 ... Answer
Q2)b
- Two independent events are defined by their probabilities as follows:
p ( A ) = 0.3 and p ( B ) = 0.5
- The occurrences of either event does not change alter or affect the occurrences of the other event; hence, independent.
- For the two events to occur simultaneously at the same time:
p ( A & B ) = p ( A )* p ( B )
p ( A & B ) = 0.3*0.5
p ( A & B ) = 0.15 ... Answer
- For either of the events to occur but not both. From the comparatively law of two independent events A and B we have:
p ( A U B ) = p ( A ) + p ( B ) - 2*p ( A & B )
p ( A U B ) = 0.3 + 0.5 - 2*0.15
p ( A U B ) = 0.5 ... Answer
- Two mutually exclusive events can-not occur simultaneously; hence, the two events are not mutually exclusive because:
p ( A & B ) = 0.15 ≠ 0
Q2)c
- The letters of the word given are to be arranged in number of different ways as follows:
STATISTICS
- Number of each letters:
S : 3
T : 3
A: 1
I: 2
C: 1
- 10 letters can be arranged in 10! ways.
- However, the letters ( S and T and I ) are repeated. So the number of permutations must be discounted by the number of each letter is repeated as follows:
- So the total number of ways the word " STATISTICS " can be re-arranged is 50,400 without repetitions.
See explanation
Step-by-step explanation:
Q1)a
- Denote a random variable ( X ) as the life time of a brand of bulb produced.
- The given mean ( μ ) = 210 hrs and standard deviation ( σ ) = 56 hrs. The distribution is symbolized as follows:
X ~ Norm ( 210 , 56^2 )
i) The bulb picked to have a life time of at least 300 hours.
- We will first standardize the limiting value of the RV ( X ) and determine the corresponding Z-score value:
P ( X ≥ x ) = P ( Z ≥ ( x - μ ) / σ )
P ( X ≥ 300 ) = P ( Z ≥ ( 300 - 210 ) / 56 )
P ( X ≥ 300 ) = P ( Z ≥ 1.607 )
- Use the standard normal look-up table for limiting value of Z-score:
P ( X ≥ 300 ) = P ( Z ≥ 1.607 ) = 0.054 .. Answer
ii) The bulb picked to have a life time of at most 100 hours.
- We will first standardize the limiting value of the RV ( X ) and determine the corresponding Z-score value:
P ( X ≤ x ) = P ( Z ≤ ( x - μ ) / σ )
P ( X ≤ 100 ) = P ( Z ≤ ( 100 - 210 ) / 56 )
P ( X ≤ 100 ) = P ( Z ≤ -1.9643 )
- Use the standard normal look-up table for limiting value of Z-score:
P ( X ≤ 100 ) = P ( Z ≤ -1.9643 ) = 0.0247 .. Answer
iii) The bulb picked to have a life time of between 150 and 250 hours.
- We will first standardize the limiting value of the RV ( X ) and determine the corresponding Z-score value:
P ( x1 ≤ X ≤ x2 ) = P ( ( x1 - μ ) / σ ≤ Z ≤ ( x2 - μ ) / σ )
P ( 150 ≤ X ≤ 250 ) = P ( ( 150 - 210 ) / 56 ≤ Z ≤ ( 250 - 210 ) / 56 )
P ( 150 ≤ X ≤ 250 ) = P ( -1.0714 ≤ Z ≤ 0.71428 )
- Use the standard normal look-up table for limiting value of Z-score:
P ( 150 ≤ X ≤ 250 ) = P ( -1.0714 ≤ Z ≤ 0.71428 ) = 0.6205 .. Answer
Q1)b
- Denote event (A) : Kofi solves the problem correctly. Then the probability of him answering successfully is:
p ( A ) = 0.25
- Denote event (B) : Menesh solves the problem correctly. Then the probability of him answering successfully is:
p ( B ) = 0.4
- The probability that neither of them answer the question correctly is defined by a combination of both events ( A & B ). The two events are independent.
- So for independent events the required probability can be stated as:
p ( A' & B' ) = p ( A' ) * p ( B' )
p ( A' & B' ) = [ 1 - p ( A ) ] * [ 1 - p ( B ) ]
p ( A' & B' ) = [ 1 - 0.25 ] * [ 1 - 0.4 ]
p ( A' & B' ) = 0.45 ... Answer
Q2)a
- A discrete random variable X: defines the probability of getting each number on a biased die.
- From the law of total occurrences. The sum of probability of all possible outcomes is always equal to 1.
∑ p ( X = xi ) = 1
p ( X = 1 ) + p ( X = 2 ) + p ( X = 3 ) + p ( X = 4 ) + p ( X = 5 ) + p ( X = 6 )
1/6 + 1/6 + 1/5 + k + 1/5 + 1/6 = 1
k = 0.1 ... Answer
- The expected value E ( X ) or mean value for the discrete distribution is determined from the following formula:
E ( X ) = ∑ p ( X = xi ) . xi
E ( X ) = (1/6)*1 + (1/6)*2 + (1/5)*3 + (0.1)*4 + (1/5)*5 + (1/6)*6
E ( X ) = 3.5 .. Answer
- The expected-square value E ( X^2 ) or squared-mean value for the discrete distribution is determined from the following formula:
E ( X^2 ) = ∑ p ( X = xi ) . xi^2
E ( X^2 ) = (1/6)*1 + (1/6)*4 + (1/5)*9 + (0.1)*16 + (1/5)*25 + (1/6)*36
E ( X^2 ) = 15.233 .. Answer
- The variance of the discrete random distribution for the variable X can be determined from:
Var ( X ) = E ( X^2 ) - [ E ( X ) ] ^2
Var ( X ) = 15.2333 - [ 3.5 ] ^2
Var ( X ) = 2.9833 ... Answer
- The cumulative probability of getting any number between 1 and 5 can be determined from the sum:
P ( 1 < X < 5 ) = P ( X = 2 ) + P ( X = 3 ) + P ( X = 4 )
P ( 1 < X < 5 ) = 1/6 + 1/5 + 0.1
P ( 1 < X < 5 ) = 0.467 ... Answer
Q2)b
- Two independent events are defined by their probabilities as follows:
p ( A ) = 0.3 and p ( B ) = 0.5
- The occurrences of either event does not change alter or affect the occurrences of the other event; hence, independent.
- For the two events to occur simultaneously at the same time:
p ( A & B ) = p ( A )* p ( B )
p ( A & B ) = 0.3*0.5
p ( A & B ) = 0.15 ... Answer
- For either of the events to occur but not both. From the comparatively law of two independent events A and B we have:
p ( A U B ) = p ( A ) + p ( B ) - 2*p ( A & B )
p ( A U B ) = 0.3 + 0.5 - 2*0.15
p ( A U B ) = 0.5 ... Answer
- Two mutually exclusive events can-not occur simultaneously; hence, the two events are not mutually exclusive because:
p ( A & B ) = 0.15 ≠ 0
Q2)c
- The letters of the word given are to be arranged in number of different ways as follows:
STATISTICS
- Number of each letters:
S : 3
T : 3
A: 1
I: 2
C: 1
- 10 letters can be arranged in 10! ways.
- However, the letters ( S and T and I ) are repeated. So the number of permutations must be discounted by the number of each letter is repeated as follows:
- So the total number of ways the word " STATISTICS " can be re-arranged is 50,400 without repetitions.
A. H 0 μ1 = μ2 = μ3
Ha μ1 ≠ 2μ ≠ μ3
2. Test Statistics is 95%
3. Critical F-Value is 3.76.
4. P-Value is 2.32.
5. Conclusion Reject the null hypothesis.
6. Type of vehicle does effect the amount of green house gas emissions.
The correct order of the steps of a hypothesis test is given below.
1. Determine the null and alternative hypothesis.
2. Select a sample and compute the critical value F-test for the sample mean.
3. Determine the probability at which you will conclude that the sample outcome is very unlikely.
4. Make a decision about the unknown population.
All steps are performed in the given sequence to test a hypothesis.
The null hypothesis is rejected or accepted on the basis of level of significance. When the p-value is greater than level of significance we fail to reject the null hypothesis and null hypothesis is then accepted. It is not necessary that all null hypothesis will be rejected at 95% level of significance. To determine the criteria for accepting or rejecting a null hypothesis we should also consider p-value.
For more details on hypothesis test follow the link:
link
A. H 0 μ1 = μ2 = μ3
Ha μ1 ≠ 2μ ≠ μ3
2. Test Statistics is 95%
3. Critical F-Value is 3.76.
4. P-Value is 2.32.
5. Conclusion Reject the null hypothesis.
6. Type of vehicle does effect the amount of green house gas emissions.
The correct order of the steps of a hypothesis test is given below.
1. Determine the null and alternative hypothesis.
2. Select a sample and compute the critical value F-test for the sample mean.
3. Determine the probability at which you will conclude that the sample outcome is very unlikely.
4. Make a decision about the unknown population.
All steps are performed in the given sequence to test a hypothesis.
The null hypothesis is rejected or accepted on the basis of level of significance. When the p-value is greater than level of significance we fail to reject the null hypothesis and null hypothesis is then accepted. It is not necessary that all null hypothesis will be rejected at 95% level of significance. To determine the criteria for accepting or rejecting a null hypothesis we should also consider p-value.
For more details on hypothesis test follow the link:
link
The test is not significant at 5% level of significance, hence we conclude that there's no variation among the discussion sections.
Step-by-step explanation:
Assumptions:
1. The sampling from the different discussion sections was independent and random.
2. The populations are normal with means and constant variance
There's no variation among the discussion sections
There's variation among the discussion sections
Df Sum Sq Mean sq F value Pr(>F)
Section 7 525.01 75 1.87 0.99986
Residuals 189 7584.11 40.13
Test Statistic =
Since our p-value is greater than our level of significance (0.05), we do not reject the null hypothesis and conclude that there's no significant variation among the eight discussion sections.
The test is not significant at 5% level of significance, hence we conclude that there's no variation among the discussion sections.
Step-by-step explanation:
Assumptions:
1. The sampling from the different discussion sections was independent and random.
2. The populations are normal with means and constant variance
There's no variation among the discussion sections
There's variation among the discussion sections
Df Sum Sq Mean sq F value Pr(>F)
Section 7 525.01 75 1.87 0.99986
Residuals 189 7584.11 40.13
Test Statistic =
Since our p-value is greater than our level of significance (0.05), we do not reject the null hypothesis and conclude that there's no significant variation among the eight discussion sections.
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