Answer: 0.9896
Step-by-step explanation:
Given : A company produces steel rods. The lengths of the steel rods are normally distributed with
Mean : [tex]\mu=208.6\text{ -cm}[/tex]
Standard deviation : [tex]\sigma=0.6\text{ -cm}[/tex]
Sample size : = 12
Let x be the random variable that represents the lengths of the steel rods .
To find probability , first we find z-score
Z-score : [tex]z=\dfrac{x-\mu}{\dfrac{\sigma}{\sqrt{n}}}[/tex]
For x= 208.2-cm.
[tex]z=\dfrac{208.2-208.6}{\dfrac{0.6}{\sqrt{12}}}\approx-2.31[/tex]
By using standard normal distribution table , the probability that the mean length of a randomly selected bundle of steel rods is greater than 208.2-cm :-
[tex]P(X>208.2)=P(z>-2.31)=1-P(z<2.01)\\\\=1-0.0104441=0.9895559\appprox0.9896[/tex]
Hence, the required probability is 0.9896.
