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Based on historical data, your manager believes that 33% of the company's orders come from first-time customers. A random sample of 163 orders will be used to estimate the proportion of first-time-customers. What is the probability that the sample proportion is greater than 0.28

Respuesta :

MrRoyal

Answer:

[tex]P(x > 0.28) = 1[/tex]

Step-by-step explanation:

Given

[tex]p = 33\%[/tex] --- orders from first time customer

[tex]n =163[/tex] --- samples

Required

[tex]P(x >0.28)[/tex]

First, calculate the mean

[tex]\bar x = np[/tex]

[tex]\bar x = 163 * 33\%[/tex]

[tex]\bar x = 53.79[/tex]

Next, the standard deviation

[tex]\sigma = \sqrt{\bar x * (1 - p)[/tex]

[tex]\sigma = \sqrt{53.79* (1 - 33\%)[/tex]

[tex]\sigma = \sqrt{53.79* (1 - 0.33)[/tex]

[tex]\sigma = \sqrt{53.79* 0.67[/tex]

[tex]\sigma = 6.00[/tex]

For [tex]P(x >0.28)[/tex],

The z score is:

[tex]z = \frac{x - \bar x}{\sigma}[/tex]

[tex]z = \frac{0.28 - 53.79}{6}[/tex]

[tex]z = \frac{-53.51}{6}[/tex]

[tex]z = -8.92[/tex]

So:

[tex]P(x > 0.28) = P(z > -8.92)[/tex]

From z probability:

[tex]P(z > -8.92) =1[/tex]

Hence:

[tex]P(x > 0.28) = 1[/tex]

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