geduiwelh

2021-02-08

A population of values has a normal distribution with

Find the probability that a single randomly selected value is less than 191.2.

Find the probability that a sample of size

Write your answers as numbers accurate to 4 decimal places.

l1koV

Skilled2021-02-09Added 100 answers

Solution for Step 1: Make X a random variable.

Using the information provided, X exhibits a normal distribution with mean $\mu =204.3$ and a standard deviation is $\sigma =43$.

Step 2 Next, determine the likelihood that a single randomly chosen figure will be less than 191.2.

$P\left(X<191.2\right)=P\left(\frac{X-\mu}{\sigma}<\frac{191.2-204.3}{43}\right)$

$=P(Z<-0.305)$

$=0.3802[Using\text{}the\text{}excel\text{}function=NORM.DIST(-0.305,0,1,TRUE)]$

As a result, 0.3802 percent chance exists that a single randomly chosen figure will be lower than 191.2.

Step 3

A sample size is determined from the provided data $n=111$.

Consequently, the likelihood that a randomly chosen sample has a mean below 191.2

$P\left(M<191.2\right)=P\left(\frac{M-\mu}{\frac{\sigma}{\sqrt{n}}}<\frac{191.2-204.3}{\frac{43}{\sqrt{111}}}\right)$

$=P\left(Z<\frac{\sqrt{111}(-13.1)}{43}\right)$

$=P(Z<-3.210)$

$=0.0007[Using\text{}the\text{}excel\text{}function\text{}=NORM.DIST(-3.210,0,1,TRUE)]$

Therefore, there is a 0.0007 percent chance that a randomly chosen sample has a mean below 191.2.

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