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1. Decide if it is appropriate to use the normal distribution to approximate the random variable x for a binomial experiment with sample size of n = 6 and probability of success p = 0.1.
2. Ten percent of the population is left-handed. A class of 100 students is selected. Convert the binomial probability P(x > 12) to a normal probability by using the correction for continuity.
a. P(x ≥ 11.5)
b. P(x > 12.5)
c. P(x < 11.5)
d. P(x ≤ 12.5)

3. An airline reports that it has been experiencing a 15% rate of no-shows on advanced reservations. Among 150 advanced reservations, find the probability that there will be fewer than 20 no-shows.
a. 0.3187
b. 0.7549
c. 0.7967
d. 0.2451

4. In a recent survey, 83% of the community favored building a police substation in their neighborhood. You randomly select 18 citizens and ask each if he or she thinks the community needs a police substation.
Decide whether you can use the normal distribution to approximate the binomial distribution. If so, find the mean and standard deviation. If not, explain why.

Respuesta :

semsee45

Answer:

1)  Not appropriate.

2)  b. P(x > 12.5)

3)  d. 0.2451

4)  Not appropriate.

Step-by-step explanation:

Question 1

For the normal approximation to a binomial distribution to work well, the following conditions need to be true:

Suppose the random variable X follows a binomial distribution:

  • X ~ B(n, p)

If p ≈ 0.5 and n is large then X can be approximated by the normal random variable:

  • Y ~ N(np, np(1 - p))

However, even if p isn’t all that close to 0.5, this approximation usually works well as long as np and n(1 – p) are both bigger than 5.

Given:

  • p = 0.1
  • n = 6

⇒ np = 6 × 0.1 = 0.6 < 5

⇒ n(1 - p) = 6 × 0.9 = 5.4 > 5

As p = 0.1 is not close to 0.5 and n is small, and np and n(1 – p) are not both bigger than 5, it is not appropriate to use the normal distribution to approximate the random variable x for a binomial experiment.

Question 2

The binomial distribution is discrete, but a normally-distributed variable is continuous.  Therefore, to allow for this, use a continuity correction.

The interval you need to use with the normal distribution depends on the discrete probability you’re trying to find, but the general idea is always the same:  

  • Each discrete value b covers the continuous interval from b - 0.5 up to b + 0.5.

Therefore, to approximate the discrete probability P(x > 12), use the continuity correction P(X > b) ≈ P(Y > b+0.5):

  • P(x > 12) ≈ P(Y > 12.5)

Question 3

There are a fixed number of trials (150 advanced reservations), with probability of success (i.e. no-shows) 0.15.  If X is the number of no-shows on advanced reservations, then X ~ B(150, 0.15).

As n is large and np = 22.5 > 5 and n(1 – p) = 127.5 > 5, a normal approximation is appropriate.

X can be approximated by a normal random variable Y ~ N(μ, σ²):

    ⇒ μ = np = 150 × 0.15 = 22.5

    ⇒ σ² = np(1 - p) = 150 × 0.15 × 0.85 = 19.125

So Y ~ N(22.5, 19.125).

Use the approximation to estimate the probability that there will be fewer than 20 no-shows.

Using the continuity correction, P(X < b) ≈ P(Y < b-0.5):

  • P(X < 20) ≈ P(Y < 19.5) = 0.2464

The probability using the binomial distribution X ~ B(150, 0.15) is:

  • P(X < 20) = P(X ≤ 19) = 0.2509

As 0.2464 ≈ 0.2509, this further proves that it is appropriate to use the normal distribution to approximate.

Question 4

There are a fixed number of trials (18 citizens), with probability of success (i.e. favored building a police substation) 0.83.  If X is the number of citizens asked, then X ~ B(18, 0.83).

Calculate np and n(1 – p):

  • np = 18 × 0.83 = 14.94 > 5
  • n(1 – p) = 18 × 0.17 = 3.06 < 5

As n is not large and p is not close to 0.5, and np and n(1 – p) are not both bigger than 5, a normal approximation is not appropriate.

Enaji6542

Answer:

semsee45  is right.

Step-by-step explanation:

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