Question

The ages of rocks in an isolated area are known to be approximately normally distributed with a mean of 7 years and a standard deviation of 1.8 years. What percent of rocks are between 5.8 and 7.3 years old?

Answer 1

Answer: 31.39%

Explanation:

Given: The ages of rocks in an isolated area are known to be approximately normally distributed with a mean of 7 years and a standard deviation of 1.8 years.

i.e.
`\mu = 7 \ \ \ \ \sigma= 1.8`

Let X be the age of rocks in an isolated area .

Then, the probability that rocks are between 5.8 and 7.3 years old :

`P(5.8<X<7.3)=P((5.8-7)/(1.8)<(X-\mu)/(\sigma)<(7.3-7)/(1.8))\\\\=P(-0.667<Z<0.167)\ \ \ [z=(X-\mu)/(\sigma)]\\\\=P(Z<0.167)-P(z<-0.667)\\\\=P(Z<0.167)-(1-P(z<0.667))\\\\=0.5663-(1-0.7476)\\\\=0.3139`

`=31.39\%`

Hence, the percent of rocks are between 5.8 and 7.3 years old = 31.39%