(1 point) In 2012, the percentage of cell phone owners who used their cell phone to send or receive text messages was 80%80%. A polling firm contacted a simple random sample of 1000 people chosen from the population of cell phone owners. If pp is the proportion of cell phone owners who used their phone to text, then what was the the probability that p^p^ was between 0.76 and 0.84?

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08-11-2019
Mathematics

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(1 point) In 2012, the percentage of cell phone owners who used their cell phone to send or receive text messages was 80%80%. A polling firm contacted a simple random sample of 1000 people chosen from the population of cell phone owners. If pp is the proportion of cell phone owners who used their phone to text, then what was the the probability that p^p^ was between 0.76 and 0.84?

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JeanaShupp JeanaShupp · Answer 1 · 2019-11-11T09:10:52+01:00

Answer: 0.9984224

Step-by-step explanation:

Let p is the proportion of cell phone owners who used their phone to text.

Given : The probability of cell phone owners who used their cell phone to send or receive text messages was : p= 0.80

sample size : n= 1000

Then, the the probability that [tex]\hat{p}[/tex] was between 0.76 and 0.84 will be :-

[tex]P(0.76<p<0.84)=P(\dfrac{0.76-0.80}{\sqrt{\dfrac{0.80(0.20)}{1000}}}<\dfrac{\hat{p}-p}{\sqrt{\dfrac{p(1-p)}{n}}}< \dfrac{0.84-0.80}{\sqrt{\dfrac{0.80(0.20)}{1000}}} )\\\\=P(-3.16<z< 3.16)=1-2P(z>3.16)\ \ [\because P(-z<Z<z)=1-2P(Z>|z|)]\\\\=1-2(0.0007888)=0.9984224[/tex]

Hence, the required probability = 0.9984224