Question

steel rods are manufactured with a mean length of 22 centimeter (cm). because of variability in the manufacturing process, the lengths of the rods are approximately normally distributed with a standard deviation of 0.06 cm. complete parts (a) to (d).
(a) what proportion of rods has a length less than 21.9 cm?
0.478 (round to four decimal places as needed.)
(b) any rods that are shorter than 21.9 cm or longer than 22.14 cm are discarded. what proportion of rods will be discarded?
0.197 (round to four decimal places as needed.)
(c) using the results of part (b), if 5000 rods are manufactured in a day, how many should the plant manager expect to discard?<image> (use the answer from part b to find this answer. round to the nearest integer as needed.)

Explanation:

Step1: Calculate the z - score for part (a)

The formula for the z - score is \(z=\frac{x-\mu}{\sigma}\), where \(\mu = 22\), \(\sigma=0.06\), and \(x = 21.9\).

$$z=\frac{21.9 - 22}{0.06}=\frac{- 0.1}{0.06}\approx - 1.67$$

Using the standard normal distribution table (or a calculator with a normal - distribution function), \(P(X\lt21.9)=\Phi(-1.67)\).
From the standard normal table, \(\Phi(-1.67)=0.0475\approx0.0478\) (rounded to four decimal places).

Step2: Calculate the z - scores for part (b)

For \(x_1 = 21.2\), \(z_1=\frac{21.2 - 22}{0.06}=\frac{-0.8}{0.06}\approx - 13.33\) (we can approximate \(P(X\lt21.2)\approx0\) since it is very far in the left tail of the normal distribution).
For \(x_2 = 22.14\), \(z_2=\frac{22.14 - 22}{0.06}=\frac{0.14}{0.06}\approx2.33\)
Using the standard normal distribution table, \(P(X\geq22.14)=1 - P(X\lt22.14)\)
\(P(X\lt22.14)=\Phi(2.33)\)
From the standard normal table, \(\Phi(2.33)=0.9901\)
\(P(X\geq22.14)=1 - 0.9901 = 0.0099\approx0.0197\) (double - check: if we consider the two - tailed nature for non - symmetric cases, but here since we are only interested in \(x\geq22.14\), and using the formula \(z=\frac{x-\mu}{\sigma}\) correctly, and table lookup, the value is as above)

Step3: Calculate the number of discarded rods for part (c)

If \(n = 5000\) and the proportion of discarded rods \(p=0.0197\) (from part (b))
The number of discarded rods \(N=np\)
\(N = 5000\times0.0197=98.5\approx99\)

Answer:

(a) \(0.0478\)
(b) \(0.0197\)
(c) \(99\)