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. Find the probability that the average length of rods in a randomly selected bundle of steel rods is greater than 259-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:

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.

1 x=22.52

2: 65/39=604+39x

65x-39x=604

26x=604

x=604/26

×=23.23