 # A population of values has a normal distribution with \mu=204.3 and \sigma=43. You intend to draw a random sample of size n=111.Find the probability that geduiwelh 2021-02-08 Answered

A population of values has a normal distribution with $\mu =204.3$ and $\sigma =43$. You intend to draw a random sample of size $n=111$.
Find the probability that a single randomly selected value is less than 191.2.
$P\left(X<191.2\right)=$?
Find the probability that a sample of size $n=111$ is randomly selected with a mean less than 191.2.
$P\left(M<191.2\right)=$?

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Step 1
Solution:
Let X be random variable.
From the given information, X follows normal distribution with mean $\mu =204.3$ and a standard deviation is $\sigma =43$.
Step 2
Then, the probability that a single randomly selected value is less than 191.2 is
$P\left(X<191.2\right)=P\left(\frac{X-\mu }{\sigma }<\frac{191.2-204.3}{43}\right)$
$=P\left(Z<-0.305\right)$

Thus, the probability that a single randomly selected value is less than 191.2 is 0.3802.
Step 3
From the given information, a sample size is $n=111$.
Then, the probability that a randomly selected with a mean less than 191.2 is
$P\left(M<191.2\right)=P\left(\frac{M-\mu }{\frac{\sigma }{\sqrt{n}}}<\frac{191.2-204.3}{\frac{43}{\sqrt{111}}}\right)$
$=P\left(Z<\frac{\sqrt{111}\left(-13.1\right)}{43}\right)$
$=P\left(Z<-3.210\right)$

Thus, the probability that a randomly selected with a mean less than 191.2 is 0.0007.