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mars1129 [50]
3 years ago
11

Suppose that weekly income of migrant workers doing agricultural labor in Florida has a distribution with a mean of $520 and a s

tandard deviation of $90. A researcher randomly selected a sample of 100 migrant workers. What is the probability that sample mean is less than $510
Mathematics
1 answer:
Finger [1]3 years ago
8 0

Answer:

z=\frac{510-520}{\frac{90}{\sqrt{100}}}= -1.11

And we can find the probability using the normal standard distribution table and with the complement rule we got:

P(z

Step-by-step explanation:

For this problem we have the following parameters:

\mu = 520, \sigma = 90

We select a sample size of n =100 and we want to find this probability:

P(\bar X

The distribution for the sample mean using the central limit theorem would be given by:

\bar X \sim N(\mu,\frac{\sigma}{\sqrt{n}})

And we can solve this problem with the z score formula given by:

z=\frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}

And if we find the z score formula we got:

z=\frac{510-520}{\frac{90}{\sqrt{100}}}= -1.11

And we can find the probability using the normal standard distribution table and with the complement rule we got:

P(z

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