A company produces steel rods. The lengths of the steel rods are normally distributed with a mean of 116.4-cm and a standard deviation of 0.8-cm. For shipment, 29 steel rods are bundled together. Find the probability that the average length of a randomly selected bundle of steel rods is less than 116-cm.

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:  

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  = 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  =    ~ N(0,1)

where,  = population mean length of rods = 259.2 cm

           = 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( > 259 cm)

    P( > 259 cm) = P(  >  ) = 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.

Calculate lengths from each of the lengths (116.4 & 0.8), using the well known formula z = (x-μ) / σ.

Look up the Probabilities associated with those two lengths

Subtract the lesser probability from greater of these two probabilities, to get probability 0.8 in the associated interval.  

it 2

step-by-step explanation:

8^1/3=2

step-by-step explanation: