Answer:
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 ) Â
- 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 ) Â
- 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 ) Â
- 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. Â
- 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:
            [tex]\frac{10!}{3!3!2!} = \frac{3628800}{72} = 50,400[/tex]
- So the total number of ways the word " STATISTICS " can be re-arranged is 50,400 without repetitions.
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