Business Weekly conducted a survey of recent graduates from the top MBA programs. On the basis of the survey, assume that 60% of the recent graduates annual salary exceeds $40000. Suppose you take a simple random sample of 51 recent graduates. Note: round all z scores to decimal places and all other numbers to 4 decimal places. Find the probability that less than 70% of the 51 recent graduates have an annual salary exceeding $40000

Respuesta :

Using the normal distribution and the central limit theorem, it is found that there is a 0.9279 = 92.79% probability that less than 70% of the 51 recent graduates have an annual salary exceeding $40000.

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, we have that:

  • 60% of the recent graduates annual salary exceeds $40000, hence p = 0.6.
  • A sample of 51 is taken, hence n = 51.

The mean and the standard error are given by:

[tex]\mu = p = 0.6[/tex]

[tex]s = \sqrt{\frac{p(1 - p)}{n}} = \sqrt{\frac{0.6(0.4)}{51}} = 0.0686[/tex]

The probability that less than 70% of the 51 recent graduates have an annual salary exceeding $40000 is 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 - 0.6}{0.0686}[/tex]

[tex]Z = 1.46[/tex]

[tex]Z = 1.46[/tex] has a p-value of 0.9279.

0.9279 = 92.79% probability that less than 70% of the 51 recent graduates have an annual salary exceeding $40000.

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

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