Answer: 31.39%
Step-by-step 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. [tex]\mu = 7 \ \ \ \ \sigma= 1.8[/tex]
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 :
[tex]P(5.8<X<7.3)=P(\dfrac{5.8-7}{1.8}<\dfrac{X-\mu}{\sigma}<\dfrac{7.3-7}{1.8})\\\\=P(-0.667<Z<0.167)\ \ \ [z=\dfrac{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[/tex]
[tex]=31.39\%[/tex]
Hence, the percent of rocks are between 5.8 and 7.3 years old = 31.39%
