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State two properties of a binomial distribution.

Mathematics

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4 parts
ii
19
a
0.9452
b
0.9426
c
0.5

Step-by-step solution

11 steps

Part i

State two properties of a binomial distribution.

  1. There is a fixed number of trials, denoted by nn.
  2. Each trial has only two possible outcomes, typically labeled as "success" or "failure".
  3. The trials are independent of each other.
  4. The probability of success, pp, is constant for every trial.

Part ii19

A discrete random variable XX follows a distribution with parameters nn and pp. Determine the value of nn given that the mean and the standard deviation of XX are both 0.95.

  1. Write down the formulas for the mean and standard deviation of a binomial distribution.

    For a binomial distribution X∼B(n,p)X \sim B(n, p):
    Mean E(X)=npE(X) = np
    Variance Var(X)=np(1−p)Var(X) = np(1-p)
    Standard deviation σ=np(1−p)\sigma = \sqrt{np(1-p)}

  2. Use the given information to set up equations.

    Given E(X)=0.95E(X) = 0.95 and σ=0.95\sigma = 0.95.

    np=0.95(1)np = 0.95 \quad (1) np(1−p)=0.95(2)\sqrt{np(1-p)} = 0.95 \quad (2)
  3. Solve for pp using the standard deviation equation.

    Square both sides of equation (2):

    np(1−p)=(0.95)2np(1-p) = (0.95)^2

    Substitute np=0.95np = 0.95 from equation (1) into this equation:

    0.95(1−p)=(0.95)20.95(1-p) = (0.95)^2

    Divide both sides by 0.95:

    1−p=0.951-p = 0.95 p=1−0.95p = 1 - 0.95 p=0.05p = 0.05
  4. Solve for nn using the mean equation and the value of pp.

    Substitute p=0.05p = 0.05 into equation (1):

    n(0.05)=0.95n(0.05) = 0.95 n=0.950.05n = \frac{0.95}{0.05} n=19n = 19

    The value of nn is 19\boxed{19}.

    (iii) The volume of mineral water Sup bottles is Normally distributed with a mean of 502 ml and standard deviation 2.5 ml. A bottle of mineral water is selected at random. Determine the probability that the bottle will contain

    Given: μ=502\mu = 502 ml, σ=2.5\sigma = 2.5 ml.
    To calculate probabilities for a normal distribution, we use the z-score formula: Z=X−μσZ = \frac{X - \mu}{\sigma}.

Part a0.9452

less than 506 ml.

  1. Calculate the z-score for X=506X = 506 ml.

    Z=506−5022.5Z = \frac{506 - 502}{2.5} Z=42.5Z = \frac{4}{2.5} Z=1.6Z = 1.6
  2. Find the probability P(X<506)P(X < 506).

    P(X<506)=P(Z<1.6)P(X < 506) = P(Z < 1.6)

    Using a standard normal distribution table, P(Z<1.6)=0.9452P(Z < 1.6) = 0.9452.
    The probability is 0.9452\boxed{0.9452}.

Part b0.9426

between 495 ml and 506 ml.

  1. Calculate the z-score for X=495X = 495 ml.

    Z1=495−5022.5Z_1 = \frac{495 - 502}{2.5} Z1=−72.5Z_1 = \frac{-7}{2.5} Z1=−2.8Z_1 = -2.8
  2. Calculate the z-score for X=506X = 506 ml.

    Z2=506−5022.5Z_2 = \frac{506 - 502}{2.5} Z2=42.5Z_2 = \frac{4}{2.5} Z2=1.6Z_2 = 1.6
  3. Find the probability P(495<X<506)P(495 < X < 506).

    P(495<X<506)=P(−2.8<Z<1.6)P(495 < X < 506) = P(-2.8 < Z < 1.6) P(−2.8<Z<1.6)=P(Z<1.6)−P(Z<−2.8)P(-2.8 < Z < 1.6) = P(Z < 1.6) - P(Z < -2.8)

    From a standard normal distribution table:
    P(Z<1.6)=0.9452P(Z < 1.6) = 0.9452
    P(Z<−2.8)=P(Z>2.8)=1−P(Z<2.8)=1−0.9974=0.0026P(Z < -2.8) = P(Z > 2.8) = 1 - P(Z < 2.8) = 1 - 0.9974 = 0.0026

    P(−2.8<Z<1.6)=0.9452−0.0026P(-2.8 < Z < 1.6) = 0.9452 - 0.0026 P(−2.8<Z<1.6)=0.9426P(-2.8 < Z < 1.6) = 0.9426

    The probability is 0.9426\boxed{0.9426}.

Part c0.5

At least 502 ml

  1. Calculate the z-score for X=502X = 502 ml.

    Z=502−5022.5Z = \frac{502 - 502}{2.5} Z=02.5Z = \frac{0}{2.5} Z=0Z = 0
  2. Find the probability P(X≥502)P(X \ge 502).

    P(X≥502)=P(Z≥0)P(X \ge 502) = P(Z \ge 0)

    Since the normal distribution is symmetric about its mean (where Z=0Z=0), the probability of being greater than or equal to the mean is 0.5.

    P(Z≥0)=0.5P(Z \ge 0) = 0.5

    The probability is 0.5\boxed{0.5}.

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