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A businessman is evaluating the effectiveness of his advertising strategy in increasing sales of his products. The data shows his advertising expenditure x (in thousands of TZS) and the corresponding sales revenue y (in hundreds of thousands of TZS).

Mathematics

ScanSolve AI Answer

Answer

0.5244

Step-by-step solution

4 steps

here's the solution for question c):

c) Probability that none of the polluting trucks will be included in the inspector's sample.

This is a problem involving combinations, specifically the hypergeometric distribution, as we are selecting a sample without replacement from a finite population.

  1. Identify the parameters for the hypergeometric distribution.

    Total number of trucks (NN) = 25
    Number of polluting trucks (KK) = 3
    Number of non-polluting trucks (N−KN-K) = 25−3=2225 - 3 = 22
    Sample size (nn) = 5
    Number of polluting trucks in the sample (kk) = 0 (since none are included)

  2. Use the hypergeometric probability formula.

    The probability P(X=k)P(X=k) is given by:

    P(X=k)=(Kk)(N−Kn−k)(Nn)P(X=k) = \frac{\binom{K}{k} \binom{N-K}{n-k}}{\binom{N}{n}}

    Substitute the values:

    P(X=0)=(30)(225)(255)P(X=0) = \frac{\binom{3}{0} \binom{22}{5}}{\binom{25}{5}}

  3. Calculate the combinations.

    (30)=3!0!(3−0)!=3!1×3!=1\binom{3}{0} = \frac{3!}{0!(3-0)!} = \frac{3!}{1 \times 3!} = 1 (225)=22!5!(22−5)!=22!5!17!=22×21×20×19×185×4×3×2×1=27864\binom{22}{5} = \frac{22!}{5!(22-5)!} = \frac{22!}{5!17!} = \frac{22 \times 21 \times 20 \times 19 \times 18}{5 \times 4 \times 3 \times 2 \times 1} = 27864 (255)=25!5!(25−5)!=25!5!20!=25×24×23×22×215×4×3×2×1=53130\binom{25}{5} = \frac{25!}{5!(25-5)!} = \frac{25!}{5!20!} = \frac{25 \times 24 \times 23 \times 22 \times 21}{5 \times 4 \times 3 \times 2 \times 1} = 53130

  4. Substitute the combination values into the probability formula and calculate.

    P(X=0)=1×2786453130P(X=0) = \frac{1 \times 27864}{53130} P(X=0)=2786453130P(X=0) = \frac{27864}{53130} P(X=0)≈0.524449P(X=0) \approx 0.524449

    Rounding to four decimal places:

    P(X=0)≈0.5244P(X=0) \approx 0.5244

    The probability that none of the polluting trucks will be included in the inspector's sample is 0.5244\boxed{0.5244}.

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