Question

Smith is a weld inspector at a shipyard. He knows from keeping track of good and substandard welds that forthe afternoon shift 5% of all welds done will be substandard. If Smith checks 300 of the 7500 welds completedthat shift, what is the probability that he will find less than 20 substandard welds

Answer 1

Answer:

0.9066

Step-by-step explanation:

According to the question, the data provided in the question is as follows

Probability =  p = 0.05

Sample size = n = 300

Total welds = 7,500

Based on the above information, the probability for less than 20 substandard welds is

As per binomial distribution, the normal approximation is

$\mu = n \times p$

$= 300 \times 0.05$

= 15

$\sigma = \sqrt{n\times p\times(1 - p)}$

$=\sqrt{300\times0.05\times(1 - 0.05)}$

= 3.77

$z = \frac{x - \mu}{\sigma}$

$= \frac{20-15}{3.77}$

= 1.32

Therefore, the probability is 0.9066