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

`P(370.2<X<450.2)=P((370.2-\mu)/(\sigma)<(X-\mu)/(\sigma)<(450.2-\mu)/(\sigma))=P((370.2-420.2)/(105)<Z<(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`

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=(x-\mu)/(\sigma)`

And replacing we got:

`P(370.2<X<450.2)=P((370.2-\mu)/(\sigma)<(X-\mu)/(\sigma)<(450.2-\mu)/(\sigma))=P((370.2-420.2)/(105)<Z<(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`