Step 1
According to the provided information,
\(\displaystyle\mu={29.3}\)
\(\displaystyle\sigma={65.1}\)
\(\displaystyle{n}={142}\)
\(\displaystyle{P}{\left({27.7}{<}{X}{<}{35.3}\right)}={P}{\left({\frac{{{27.7}-{29.3}}}{{\frac{{65.1}}{\sqrt{{{142}}}}}}}{<}{\frac{{{X}-\mu}}{{\frac{\sigma}{\sqrt{{{n}}}}}}}{<}{\frac{{{35.3}-{29.3}}}{{\frac{{65.1}}{\sqrt{{{142}}}}}}}\right)}\)
\(\displaystyle={P}{\left(-{0.29}{<}{Z}{<}{1.1}\right)}{\left({Z}={\frac{{{X}-\mu}}{{\sigma}}}\right)}\)
\(\displaystyle={0.4784}\) (From standard normal table)
Therefore, the probability that a sample of size \(\displaystyle{n}={142}\) is randomly selected with a mean between 27.7 and 35.3.
\(\displaystyle{P}{\left({27.7}{<}\overline{{{X}}}{<}{35.3}\right)}={0.4784}\)
A population of values has a normal distribution with \mu=29.3 and \sigma=65.1. You intend to draw a random sample of size n=142.
A population of values has a normal distribution with \mu=29.3 and \sigma=65.1. You intend to draw a random sample of size n=142.
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Answers (1)
2020-12-01
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