Answer:
The probability that the average length of rods in a randomly selected bundle of steel rods is greater than 259 cm is 0.65173.
Step-by-step explanation:
We are given that a company produces steel rods. The lengths of the steel rods are normally distributed with a mean of 259.2 cm and a standard deviation of 2.1 cm. For shipment, 17 steel rods are bundled together.
Let [tex]\bar X[/tex] = the average length of rods in a randomly selected bundle of steel rods
The z-score probability distribution for the sample mean is given by;
Z = [tex]\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }[/tex] ~ N(0,1)
where, [tex]\mu[/tex] = population mean length of rods = 259.2 cm
[tex]\sigma[/tex] = standard deviaton = 2.1 cm
n = sample of steel rods = 17
Now, the probability that the average length of rods in a randomly selected bundle of steel rods is greater than 259 cm is given by = P([tex]\bar X[/tex] > 259 cm)
P([tex]\bar X[/tex] > 259 cm) = P( [tex]\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }[/tex] > [tex]\frac{259-259.2}{\frac{2.1}{\sqrt{17} } }[/tex] ) = P(Z > -0.39) = P(Z < 0.39)
= 0.65173
The above probability is calculated by looking at the value of x = 0.39 in the z table which has an area of 0.65173.
