Question

Smith is a weld inspector at the shipyard. He knows you're keeping track of good and substandard welds that for the afternoon shift 5% of all worlds done will be substandard is Smith checks 300 of the Seney 500 rounds completed the shift which is probably that he will find less than 20 substandard Welds

Answer 1

We are given with
5% chance that all welds will be substandard = P

500 total rounds = N
300 rounds checked = n

We asked for the probability that less than x = 20 welds will be substandard

Find the proportion
p = x / n = 25/300

Determine the standard deviation
s = √ P (1 - P) /n
s = √ 0.05(1 - 0.05) / 300

Solve for the z-score
z = (p - P)/s

Refer to the z-score table to determine the probability

Answer 2

Answer:

z = 3.87 is normal and definitely it is usual

Explanation:

given data

probability  P = 5 %

sample size: n = 300

solution

we use here Poisson Distribution and get average number of welds

average number of welds m = n × p

average number of welds m = 300 × 0.05

m = 1 5

and

standard deviation is = $\sqrt{np(1-p)}$

standard deviation is = $\sqrt{300*0.05*0.95}$

standard deviation is = 3.77

so here by normal distribution

z value corresponding 30 sub standard

so z will be

z = x - Mean ÷ standard deviation  

z = $\frac{20-15}{3.77}$  

z = 1.326

as z value base on standard normal curves that have maximum value = 3.77

z value = 1.32

so z = 3.87 is normal and definitely it is usual