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Given that X is a binomial distribution with n=8 and p=0.6. The probability mass function is P(X=k) = nk p^k (1-p)ⁿ-k.

Computer Science

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Answers

3 parts
a

0.2787\text{0.2787}

b

0.1064\text{0.1064}

c

0.6348\text{0.6348}

Step-by-step solution

16 steps

Given that XX is a binomial distribution with n=8n=8 and p=0.6p=0.6. The probability mass function is P(X=k)=(nk)pk(1−p)n−kP(X=k) = \binom{n}{k} p^k (1-p)^{n-k}.
Here, n=8n=8, p=0.6p=0.6, and 1−p=0.41-p = 0.4.

Part a· 5 steps

Find P(X=5)P(X=5)

  1. Apply the binomial probability formula.

    P(X=5)=(85)(0.6)5(0.4)8−5P(X=5) = \binom{8}{5} (0.6)^5 (0.4)^{8-5} P(X=5)=(85)(0.6)5(0.4)3P(X=5) = \binom{8}{5} (0.6)^5 (0.4)^3
  2. Calculate the binomial coefficient.

    (85)=8!5!(8−5)!=8!5!3!=8×7×63×2×1=56\binom{8}{5} = \frac{8!}{5!(8-5)!} = \frac{8!}{5!3!} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56
  3. Calculate the powers of pp and (1−p)(1-p).

    (0.6)5=0.07776(0.6)^5 = 0.07776 (0.4)3=0.064(0.4)^3 = 0.064
  4. Multiply the values.

    P(X=5)=56×0.07776×0.064=0.27869184P(X=5) = 56 \times 0.07776 \times 0.064 = 0.27869184
  5. Round to 4 decimal places.

    P(X=5)≈0.2787P(X=5) \approx 0.2787

    The final answer is 0.2787\boxed{\text{0.2787}}

Part b

0.1064\text{0.1064}

Find P(X>6)P(X > 6)

  1. Express P(X>6)P(X > 6) as a sum of probabilities.

    P(X>6)=P(X=7)+P(X=8)P(X > 6) = P(X=7) + P(X=8)
  2. Calculate P(X=7)P(X=7).

    P(X=7)=(87)(0.6)7(0.4)8−7=(87)(0.6)7(0.4)1P(X=7) = \binom{8}{7} (0.6)^7 (0.4)^{8-7} = \binom{8}{7} (0.6)^7 (0.4)^1 (87)=8\binom{8}{7} = 8 (0.6)7=0.0279936(0.6)^7 = 0.0279936 P(X=7)=8×0.0279936×0.4=0.08957952P(X=7) = 8 \times 0.0279936 \times 0.4 = 0.08957952
  3. Calculate P(X=8)P(X=8).

    P(X=8)=(88)(0.6)8(0.4)8−8=(88)(0.6)8(0.4)0P(X=8) = \binom{8}{8} (0.6)^8 (0.4)^{8-8} = \binom{8}{8} (0.6)^8 (0.4)^0 (88)=1\binom{8}{8} = 1 (0.6)8=0.01679616(0.6)^8 = 0.01679616 P(X=8)=1×0.01679616×1=0.01679616P(X=8) = 1 \times 0.01679616 \times 1 = 0.01679616
  4. Sum the probabilities.

    P(X>6)=0.08957952+0.01679616=0.10637568P(X > 6) = 0.08957952 + 0.01679616 = 0.10637568
  5. Round to 4 decimal places.

    P(X>6)≈0.1064P(X > 6) \approx 0.1064

    The final answer is 0.1064\boxed{\text{0.1064}}

Part c

0.6348\text{0.6348}

Find P(2<X<6)P(2 < X < 6)

  1. Express P(2<X<6)P(2 < X < 6) as a sum of probabilities.

    P(2<X<6)=P(X=3)+P(X=4)+P(X=5)P(2 < X < 6) = P(X=3) + P(X=4) + P(X=5)
  2. Calculate P(X=3)P(X=3).

    P(X=3)=(83)(0.6)3(0.4)8−3=(83)(0.6)3(0.4)5P(X=3) = \binom{8}{3} (0.6)^3 (0.4)^{8-3} = \binom{8}{3} (0.6)^3 (0.4)^5 (83)=8!3!5!=8×7×63×2×1=56\binom{8}{3} = \frac{8!}{3!5!} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56 (0.6)3=0.216(0.6)^3 = 0.216 (0.4)5=0.01024(0.4)^5 = 0.01024 P(X=3)=56×0.216×0.01024=0.12386304P(X=3) = 56 \times 0.216 \times 0.01024 = 0.12386304
  3. Calculate P(X=4)P(X=4).

    P(X=4)=(84)(0.6)4(0.4)8−4=(84)(0.6)4(0.4)4P(X=4) = \binom{8}{4} (0.6)^4 (0.4)^{8-4} = \binom{8}{4} (0.6)^4 (0.4)^4 (84)=8!4!4!=8×7×6×54×3×2×1=70\binom{8}{4} = \frac{8!}{4!4!} = \frac{8 \times 7 \times 6 \times 5}{4 \times 3 \times 2 \times 1} = 70 (0.6)4=0.1296(0.6)^4 = 0.1296 (0.4)4=0.0256(0.4)^4 = 0.0256 P(X=4)=70×0.1296×0.0256=0.2322432P(X=4) = 70 \times 0.1296 \times 0.0256 = 0.2322432
  4. Use P(X=5)P(X=5) from part a).

    P(X=5)=0.27869184P(X=5) = 0.27869184
  5. Sum the probabilities.

    P(2<X<6)=0.12386304+0.2322432+0.27869184=0.63479808P(2 < X < 6) = 0.12386304 + 0.2322432 + 0.27869184 = 0.63479808
  6. Round to 4 decimal places.

    P(2<X<6)≈0.6348P(2 < X < 6) \approx 0.6348

    The final answer is 0.6348\boxed{\text{0.6348}}

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