Question

In a Scrabble tournament, the scores were normally distributed and the mean score was 420.2 points with a standard deviation of 105.0 points. What is the probability that the score of a randomly selected competitor differs from the mean score by less than 50 points

Answer 1

Answer:

$P(370.2<X<450.2)=P(\frac{370.2-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{450.2-\mu}{\sigma})=P(\frac{370.2-420.2}{105}<Z<\frac{450.2-420.2}{105})=P(-0.762<z<0.286)$

And we can find this probability with this difference:

$P(-0.762<z<0.286)=P(z<0.286)-P(z<-0.762)= 0.613- 0.223= 0.390$

Step-by-step explanation:

Let X the random variable that represent the scores of a population, and for this case we know the distribution for X is given by:

$X \sim N(420.2,105)$  

Where $\mu=420.2$ and $\sigma=105$

We are interested on this probability

$P(X<50)$

And we can use the z score formula given by:

$z=\frac{x-\mu}{\sigma}$

And replacing we got:

$P(370.2<X<450.2)=P(\frac{370.2-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{450.2-\mu}{\sigma})=P(\frac{370.2-420.2}{105}<Z<\frac{450.2-420.2}{105})=P(-0.762<z<0.286)$

And we can find this probability with this difference:

$P(-0.762<z<0.286)=P(z<0.286)-P(z<-0.762)= 0.613- 0.223= 0.390$