Question

A random sample of 200 screws is drawn from a population which represents the size of screws. If a sample is distributed normally with mean 3.15 cm and standard deviation 0.025cm, find expected number of screws whose size falls between 3.12 cm and 3.2 cm. [5]

[Given A(z = 1.2) = 0.3849, A(z = 2) = 0.4772]

Answer 1

The expected number of screws whose size falls between 3.12 cm and 3.2 cm is 172.42.

Answer 2

1. Identify the mean ($\mu = 3.15$ cm) and standard deviation ($\sigma = 0.025$ cm) of the normally distributed screw sizes.

2. Convert the given range of screw sizes (3.12 cm to 3.2 cm) into standard Z-scores using the formula $Z = \frac{X - \mu}{\sigma}$.

   • For $X_1 = 3.12$: $Z_1 = \frac{3.12 - 3.15}{0.025} = \frac{-0.03}{0.025} = -1.2$.
   • For $X_2 = 3.2$: $Z_2 = \frac{3.2 - 3.15}{0.025} = \frac{0.05}{0.025} = 2$.

3. Calculate the probability that a randomly selected screw falls within this range by finding the area under the standard normal curve between $Z_1 = -1.2$ and $Z_2 = 2$. This probability is $P(-1.2 \le Z \le 2)$.

4. Using the symmetry of the normal distribution and the given areas $A(z) = P(0 \le Z \le z)$:

   $P(-1.2 \le Z \le 2) = P(-1.2 \le Z \le 0) + P(0 \le Z \le 2)$

   $P(-1.2 \le Z \le 0) = P(0 \le Z \le 1.2) = A(1.2) = 0.3849$.

   $P(0 \le Z \le 2) = A(2) = 0.4772$.

   So, $P(-1.2 \le Z \le 2) = 0.3849 + 0.4772 = 0.8621$.

5. The expected number of screws in a sample of 200 that fall within this size range is the product of the sample size and the probability:

   Expected number = Sample size $\times P(3.12 \le X \le 3.2)$

   Expected number = $200 \times 0.8621 = 172.42$.