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Sagot :
Answer:
C.[tex]P(7.0039<x<7.8026)=0.1334[/tex]
D.[tex]P(7.0039<\bar{x}<7.8026)\approx 0.2942[/tex]
Step-by-step explanation:
We are given that
n=18
Mean, [tex]\mu=6.75[/tex]
Standard deviation, [tex]\sigma=2.28[/tex]
c.
[tex]P(7.0039<x<7.8026)=P(\frac{7.0039-6.75}{2.28}<\frac{x-\mu}{\sigma}<\frac{7.8026-6.75}{2.28})[/tex]
[tex]P(7.0039<x<7.8026)=P(0.11<Z<0.46)[/tex]
[tex]P(a<z<b)=P(z<b)-P(z<a)[/tex]
Using the formula
[tex]P(7.0039<x<7.8026)=P(Z<0.46)-P(Z<0.11)[/tex]
[tex]P(7.0039<x<7.8026)=0.67724-0.54380[/tex]
[tex]P(7.0039<x<7.8026)=0.1334[/tex]
D.[tex]P(7.0039<\bar{x}<7.8026)=P(\frac{7.0039-6.75}{2.28/\sqrt{18}}<\frac{x-\mu}{\frac{\sigma}{\sqrt{n}}})<\frac{7.8026-6.75}{2.28/\sqrt{18}})[/tex]
[tex]P(7.0039<\bar{x}<7.8026)=P(0.47<Z<1.96)[/tex]
[tex]P(7.0039<\bar{x}<7.8026)=P(Z<1.96)-P(Z<0.47)[/tex]
[tex]P(7.0039<\bar{x}<7.8026)=0.97500-0.68082[/tex]
[tex]P(7.0039<\bar{x}<7.8026)=0.29418\approx 0.2942[/tex]
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