A rough lumber mill cuts pine and hemlock lumber to order. Rough lumber is not cut to exact dimensions. It is always cut longer than specified. thus, if you buy an 8-foot board it will be longer than 8 feet. the boards will not be smooth, thus, it will be a rough surface. The boards vary in extra length between one and five inches. The extra lengths are a uniform distribution. You take a sample of 30 boards from the lumber mill. Determine the probability that the total extra length for the 30 boards is more than 105 inches.



Answer :

ANSWER:

The probability that the total extra length of 30 boards exceeds 105 inches is about 0.0088, or 0.88%. This indicates it is quite unlikely that the total extra length will exceed 105 inches under the given conditions.


EXPLANATION:

Step 1: Define the Distribution for One Board's Extra Length

Each board's extra length [tex]\( X \)[/tex] is uniformly distributed between 1 and 5 inches. The mean [tex](\( \mu \))[/tex] and variance [tex](\( \sigma^2 \))[/tex] of a uniform distribution ranging from [tex]\( a \)[/tex] to [tex]\( b \)[/tex] are given by:

[tex]\[ \mu = \frac{a + b}{2} \][/tex]

[tex]\[ \sigma^2 = \frac{(b - a)^2}{12} \][/tex]

For our case, [tex]\( a = 1 \)[/tex] and [tex]\( b = 5 \)[/tex], so:

[tex]\[ \mu = \frac{1 + 5}{2} = 3 \text{ inches} \][/tex]

[tex]\[ \sigma^2 = \frac{(5 - 1)^2}{12} = \frac{16}{12} = \frac{4}{3} \text{ square inches} \][/tex]

Step 2: Calculate the Sum of the Extra Lengths for 30 Boards

If you sum the extra lengths of 30 boards, the total extra length [tex]\( S \)[/tex] is the sum of 30 independent uniformly distributed random variables [tex]\( X_i \)[/tex]. By the Central Limit Theorem, for a large number of observations (30 is generally considered sufficient), [tex]\( S \)[/tex] approximates a normal distribution where:

[tex]\[ S \sim N(n\mu, n\sigma^2) \][/tex]

[tex]\[ \text{where } n = 30 \text{ (number of boards)} \][/tex]

Thus:

[tex]\[ \mu_S = 30 \times 3 = 90 \text{ inches} \][/tex]

[tex]\[ \sigma^2_S = 30 \times \frac{4}{3} = 40 \text{ square inches} \][/tex]

[tex]\[ \sigma_S = \sqrt{40} = \sqrt{40} \approx 6.32 \text{ inches} \][/tex]

Step 3: Compute the Probability [tex]\( P(S > 105) \)[/tex]

We standardize [tex]\( S \)[/tex] to convert it to a standard normal variable [tex]\( Z \):[/tex]

[tex]\[ Z = \frac{S - \mu_S}{\sigma_S} \][/tex]

[tex]\[ P(S > 105) = P\left(Z > \frac{105 - 90}{6.32}\right) \][/tex]

[tex]\[ = P(Z > 2.37) \][/tex]

Using standard normal distribution tables or a calculator:

[tex]\[ P(Z > 2.37) \][/tex] is approximately 0.0088 (based on standard normal tables).