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Novay_Z [31]
3 years ago
8

A city council is considering funding a proposal to create a new city park. The council members will fund the proposal if they c

onclude that more than 60 percent of the city residents support the proposal. A survey of 2,000 randomly selected city residents will be conducted to investigate the level of support for the proposal. Let X represent the number of city residents in the sample who support the proposal. Assume that X is a binomial random variable.
Determine the mean and the standard deviation of the random variable X, assuming that 60 percent of city henudng proosal to create a new city park.
Mathematics
1 answer:
Paha777 [63]3 years ago
3 0

Answer:

The mean of the the random variable <em>X</em> is 1200.

The standard deviation of the random variable <em>X</em> is 21.91.

Step-by-step explanation:

The random variable <em>X</em> is defined as the number of city residents in the sample who support the proposal.

The random variable <em>X</em> follows a Binomial distribution with parameters <em>n</em> = 2000 and <em>p</em> = 0.60.

But the sample selected is too large and the probability of success is close to 0.50.

So a Normal approximation to binomial can be applied to approximate the distribution of X if the following conditions are satisfied:

  1. np ≥ 10
  2. n(1 - p) ≥ 10

Check the conditions as follows:

np=2000\times 0.60=1200>10\\\\n(1-p)=2000\times (1-0.60)=800>10  

Thus, a Normal approximation to binomial can be applied.

So,  the random variable <em>X</em> can be approximate by the Normal distribution .

Compute the mean of <em>X</em> as follows:

\mu=np

  =2000\times 0.60\\=1200

The mean of the the random variable <em>X</em> is 1200.

Compute the standard deviation of <em>X</em> as follows:

\sigma=\sqrt{np(1-p)}

  =\sqrt{2000\times 0.60\times (1-0.60)}\\=\sqrt{480}\\=21.9089\\\approx 21.91

The standard deviation of the random variable <em>X</em> is 21.91.

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