Question

# A population of values has a normal distribution with \mu = 200 and \sigma = 31.9. You intend to draw a random sample of size n = 11

Random variables

A population of values has a normal distribution with $$\displaystyle\mu={200}$$ and $$\displaystyle\sigma={31.9}$$. You intend to draw a random sample of size $$\displaystyle{n}={11}$$.
Find the probability that a single randomly selected value is less than 226.9.
$$\displaystyle{P}{\left({X}{<}{226.9}\right)}=$$?

2020-10-29

The probability that a single randomly selected value is less than 226.9 is,
$$\displaystyle{P}{\left({X}{<}{226.9}\right)}={P}{\left({\frac{{{X}-\mu}}{{\sigma}}}{<}{\frac{{{226.9}-{200}}}{{{31.9}}}}\right)}$$
$$\displaystyle={P}{\left({z}{<}{\frac{{{26.9}}}{{{31.9}}}}\right)}$$
$$\displaystyle={P}{\left({z}{<}{0.843}\right)}$$
The probability of z less than 0.843 can be obtained using the excel formula “=NORM.S.DIST(0.843,TRUE)”. The probability value is 0.8004.
The required probability value is,
$$\displaystyle{P}{\left({X}{<}{226.9}\right)}={P}{\left({z}{<}{0.843}\right)}={0.8004}$$
Thus, the probability that a single randomly selected value is less than 226.9 is 0.8004.