2 years ago

Please hurry ASAP

Mathematics

2 answers:

lukranit [14]2 years ago

5 0

Answer:

It Is The First One

Step-by-step explanation:

ExtremeBDS [4]2 years ago

4 0

Answer:

Step-by-step explanation:

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Let p be the population proportion.

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$P(\hat{p}>0.11)=P(\dfrac{\hat{p}-p}{\sqrt{\dfrac{p(1-p)}{n}}}>\dfrac{0.11-0.10}{\sqrt{\dfrac{0.10(0.90)}{1100}}})\\\\=P(z>\dfrac{0.01}{0.0090453})\ \ \ [\because z=\dfrac{\hat{p}-p}{\sqrt{\dfrac{p(1-p)}{n}}} ]\\\\=P(z>1.1055)\\\\=1-P(z\leq1.055)\\\\=1-0.8543=0.1457\ \ \ [\text{using z-table}]$

Hence, the probability that the sample proportion is more than 0.11 = 0.1457

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