Suppose that 76% of americans prefer coke to pepsi. a sample of 80 was taken. what is the probability that at least seventy percent of the sample prefers coke to pepsi

Answer :

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The standard deviation of the sample proportion is:
[tex] \sqrt{ \frac{0.76(1-0.76)}{80}} =0.048[/tex]
The z-score for 0.7 is:
[tex]z= \frac{0.7-0.76}{0.048} =-1.25[/tex]
Reference to a standard normal distribution table shows that the probability that at least 70% of the sample prefers coke to pepsi is 0.8944 or 89.44%.

Using the normal distribution and the central limit theorem, it is found that there is a 0.8962 = 89.62% probability that at least seventy percent of the sample prefers coke to pepsi.

Normal Probability Distribution

In a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

  • It measures how many standard deviations the measure is from the mean.
  • After finding the z-score, we look at the z-score table and find the p-value associated with this z-score, which is the percentile of X.
  • By the Central Limit Theorem, for a proportion p in a sample of size n, the sampling distribution of sample proportion is approximately normal with mean [tex]\mu = p[/tex] and standard deviation [tex]s = \sqrt{\frac{p(1 - p)}{n}}[/tex], as long as [tex]np \geq 10[/tex] and [tex]n(1 - p) \geq 10[/tex].

In this problem:

  • 76% of Americans prefer coke to Pepsi, hence p = 0.76.
  • A sample of 80 was taken, hence n = 80.

The mean and the standard error are given by:

[tex]\mu = p = 0.76[/tex]

[tex]s = \sqrt{\frac{p(1 - p)}{n}} = \sqrt{\frac{0.76(0.24)}{80}} = 0.0477[/tex]

The  probability that at least seventy percent of the sample prefers coke to pepsi is the 1 subtracted by the p-value of Z when X = 0.7, hence:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

By the Central Limit Theorem

[tex]Z = \frac{X - \mu}{s}[/tex]

[tex]Z = \frac{0.7 - 0.76}{0.0477}[/tex]

[tex]Z = -1.26[/tex]

[tex]Z = -1.26[/tex] has a p-value of 0.1038.

1 - 0.1038 = 0.8962.

0.8962 = 89.62% probability that at least seventy percent of the sample prefers coke to pepsi.

To learn more about the normal distribution and the central limit theorem, you can take a look at https://brainly.com/question/24663213

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