Smith is a weld inspector at a shipyard. He knows from keeping track of good and substandard welds that for the afternoon shift 5% of all welds done will be substandard. If Smith checks 300 of the 7500 welds completed that shift, what is the probability that he will find less than 20 substandard welds? Where necessary, round to 4 decimal places.
The correct answer and explanation is :
To solve this problem, we can model the number of substandard welds Smith will find using the binomial distribution since each weld can either be substandard or not substandard. The binomial distribution is defined as:
[
P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}
]
where:
- ( n ) is the number of trials (in this case, the number of welds Smith checks),
- ( k ) is the number of successes (in this case, the number of substandard welds),
- ( p ) is the probability of success (in this case, the probability of a weld being substandard),
- ( \binom{n}{k} ) is the binomial coefficient.
However, because ( n ) is large (300 welds), we can approximate the binomial distribution with a normal distribution using the Central Limit Theorem. This approximation is reasonable when both ( np ) and ( n(1-p) ) are greater than 5.
Step 1: Parameters of the normal distribution
First, we need to find the mean (( \mu )) and standard deviation (( \sigma )) of the binomial distribution:
- Mean: ( \mu = n \times p = 300 \times 0.05 = 15 )
- Standard deviation: ( \sigma = \sqrt{n \times p \times (1 – p)} = \sqrt{300 \times 0.05 \times 0.95} \approx \sqrt{14.25} \approx 3.775 )
Thus, we can approximate the number of substandard welds found by a normal distribution with:
- ( \mu = 15 )
- ( \sigma \approx 3.775 )
Step 2: Standardizing the value
We are asked to find the probability that Smith will find less than 20 substandard welds, so we need to find:
[
P(X < 20)
]
We will standardize this value using the Z-score formula:
[
Z = \frac{X – \mu}{\sigma}
]
Substituting the values for ( X = 20 ), ( \mu = 15 ), and ( \sigma = 3.775 ):
[
Z = \frac{20 – 15}{3.775} = \frac{5}{3.775} \approx 1.325
]
Step 3: Using the Z-table
Now, we can use a Z-table to find the probability corresponding to a Z-score of 1.325. Looking up this value, we find that the cumulative probability for ( Z = 1.325 ) is approximately 0.9075.
Thus, the probability that Smith will find less than 20 substandard welds is 0.9075, or 90.75%.
Conclusion:
The probability that Smith will find fewer than 20 substandard welds out of the 300 he inspects is approximately 0.9075 or 90.75%. This is a fairly high probability, indicating that finding fewer than 20 substandard welds is quite likely.