A population of values has a normal distribution with mean 191.4 and standard deviation of 69.7. A random sample of size n = 153 is drawn. Find the probability that a single randomly selected value is between 188 and 206.6 round your answer to four decimal places. P=?

Question
Random variables
asked 2021-01-27
A population of values has a normal distribution with mean 191.4 and standard deviation of 69.7. A random sample of size \(\displaystyle{n}={153}\) is drawn.
Find the probability that a single randomly selected value is between 188 and 206.6 round your answer to four decimal places.
P=?

Answers (1)

2021-01-28
The probability that a single randomly selected value is between 188 and 206.6 is obtained below:
From the given information, a population of values has a normal distribution with mean 191.4 and standard deviation of 69.7 and sample of size \(\displaystyle{n}={153}\).
The required formula is given below:
\(\displaystyle{Z}={\frac{{{X}-\mu}}{{\sigma}}}\)
The required value is given below:
\(\displaystyle{P}{\left({188}\leq{X}\leq{206.6}\right)}={P}{\left({X}\leq{206.6}\right)}-{P}{\left({X}\leq{188}\right)}\)
\(\displaystyle={P}{\left({Z}\leq{\frac{{{206.6}-{191.4}}}{{{69.7}}}}\right)}-{P}{\left({Z}\leq{\frac{{{188}-{191.4}}}{{{69.7}}}}\right)}\)
\(\displaystyle={P}{\left({Z}\leq{\frac{{{15.2}}}{{{69.7}}}}\right)}-{P}{\left({Z}\leq{\frac{{-{3.4}}}{{{69.7}}}}\right)}\)
\(\displaystyle={P}{\left({Z}\leq{0.22}\right)}-{P}{\left({Z}\leq-{0.05}\right)}\)
\(\displaystyle={0.5871}-{0.4801}={0.1070}\)
Thus, the probability that a single randomly selected value is between 188 and 206.6 is 0.1070.
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