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The average number of moves a person makes in his or her lifetime is 12 and the standard deviation is 3.2. Assume that the sample is taken from a large population and the correction factor can be ignored. Use a TI-83 Plus/TI-84 Plus calculator and round to at least four decimal places. Find the probability that the mean of a sample of 36 people is less than 10.

Sagot :

Answer:

the probability ≈ 0.0001

Step-by-step explanation:

We can find the probability that the mean of a sample of 36 people is less than 10 by using Sampling Distribution for Normal Distribution (since the sample size is ≥ 30).

First, we have to find the Z-score for (X = 10) by using this formula:

[tex]\boxed{Z=\frac{x-\mu}{\sqrt{\frac{\sigma^2}{n}} } }[/tex]

where:

  • [tex]Z=\texttt{Z-score}[/tex]
  • [tex]x=\texttt{observed value}[/tex]
  • [tex]\mu=\texttt{mean}[/tex]
  • [tex]\sigma=\texttt{standard deviation}[/tex]
  • [tex]n=\texttt{sample size}[/tex]

Given:

  • [tex]x=10[/tex]
  • [tex]\mu=12[/tex]
  • [tex]\sigma=3.2[/tex]
  • [tex]n=36[/tex]

Then:

[tex]\begin{aligned}Z(X=10)&=\frac{x-\mu}{\sqrt{\frac{\sigma^2}{n}} } \\\\&=\frac{10-12}{\sqrt{\frac{3.2^2}{36} } } \\\\&=-3.75\end{aligned}[/tex]

By using the calculator, we can find that:

[tex]P(Z < -3.75)\approx\bf 0.0001[/tex]

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