Question

Baobab is a large tropical tree. The diameter of a fully grown trunk of Baobab is normally distributed with a mean of 30 feet and a standard deviation of 7 feet. What is the probability that the mean diameter of the trunks of randomly selected 15 Baobab trees is 27 to 34 feet (round off to third decimal place)

Answer 1

Answer: 0.938

Step-by-step explanation:

Given : The diameter of a fully grown trunk of Baobab is normally distributed with a mean of 30 feet and a standard deviation of 7 feet.

 

i.e. $\mu=30$    $\sigma= 7$

Sample size : n= 15

Let $\overline{x}$ represents the sample mean diameter of the trunks.

Then, the probability that the mean diameter of the trunks of randomly selected 15 Baobab trees is 27 to 34 feet will be :-

$P(27<\overline{x}<34)=P(\dfrac{27-30}{\dfrac{7}{\sqrt{15}}}<\dfrac{\overline{x}-\mu}{\dfrac{\sigma}{\sqrt{n}}}<\dfrac{34-30}{\dfrac{7}{\sqrt{15}}})\\\\=P(-1.66<z<2.21)\ \ [\because\ z=\dfrac{\overline{x}-\mu}{\dfrac{\sigma}{\sqrt{n}}}]\\\\=P(z<2.21)-P(z<-1.66)\\\\=P(z<2.21)-P(1-P(z<1.66))\ \ [\because\ P(Z<-z)=1-P(Z<z)]\\\\= 0.9864-(1- 0.9515)\ \ [\text{By z-table}]\\\\=0.9379\approx0.938$

Hence, the probability that the mean diameter of the trunks of randomly selected 15 Baobab trees is 27 to 34 feet is 0.938 .