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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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