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marysya [2.9K]
3 years ago
15

Polly's Polls asked 1850 second-year college students if they still had their original major. According to the colleges, 65% of

all second-year college students still had their original major. What is the probability that Polly's Polls' random sample will give a result within 1% of the true value? Assume the normal model applies here. You may use your calculator or reference the z-tables when working with normal models.
Mathematics
1 answer:
suter [353]3 years ago
4 0

Answer:

The probability that Polly's Sample will give a result within 1% of the value 65% is 0.6424

Step-by-step explanation:

The variable that assigns the value 1 if a person had its original major and 0 otherwise is a Bernoulli variable with paramenter 0.65. Since she asked the question to 1850 people, then the number of students that will have their original major is a Binomial random variable with parameters n = 1850, p = 0.65.

Since the sample is large enough, we can use the Central Limit Theorem to approximate that random variable to a Normal random variable, which we will denote X.

The parameters of X are determined with the mean and standard deviation of the Binomal that we are approximating. The mean is np = 1850*0.65 = 1202.5, and the standard deviation is √np(1-p) = √(1202.5*0.35) = 20.5152.

We want to know the probability that X is between 0.64*1850 = 1184 and 0.66*1850 = 1221 (that is, the percentage is between 64 and 66). In order to calculate this, we standarize X so that we can work with a standard normal random variable W ≈ N(0,1). The standarization is obtained by substracting the mean from X and dividing the result by the standard deviation, in other words

W = \frac{X-\lambda}{\sigma} = \frac{X-1202.5}{20.5152}

The values of the cummulative function of the standard normal variable W, which we will denote \phi are tabulated and they can be found in the attached file.

Now, we are ready to compute the probability that X is between 1184 and 1221. Remember that, since the standard random variable is symmetric through 0, then \phi(-z) = 1-\phi(z) for each positive value z.

P(1184 < X < 1221) = P(\frac{1184-1202.5}{20.5152} < \frac{X-1202.5}{20.5152} < \frac{1221-1202.5}{20.5152})\\ = P(-0.9018 < W < 0.9018) = \phi(0.9018) - \phi(-0.9018) = \phi(0.9018)-(1-\phi(0.9018))\\ = 2\phi(0.9018)-1 = 2*0.8212-1 = 0.6424

Therefore, the probability that Polly's Sample will give a result within 1% of the value 65% is 0.6424.

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