Answer:
a) Therefore percent of days the factory produced 20280 or fewer was 97.72%
b.) Therefore percent of days the factory produced 20320 or more was 0.13%
c) Therefore percent of days the factory produced 20160 or fewer was 15.87%
Step-by-step explanation:
i) Sandals produced per day at a certain factory last year were normally distributed
ii) mean = 20,200 sandals
iii) standard deviation = 40 sandals
a) what percent of days did the factory produce 20280 or fewer
therefore we have to find P(X < 20280) or P(z <( (20280 - 20200)/40))
therefore we have to find P(z < 80/40 ) or P(z < 2)
this is a left tailed test. therefore from z table we P(z < 2) we get P(Z < 2) = 0.9772
Therefore percent of days the factory produced 20280 or fewer was 97.72%
b)On what percent of the days last ear did the factory produce 20,320 or
more.
therefore we have to find P(X > 20320) or P(z > ( (20320 - 20200)/40))
therefore we have to find P(z > 120/40 ) or P(z > 3)
this is a right tailed test. therefore from z table we P(z > 3) we get 1 - P(Z
< 3) = 1 - 0.9987 = 0.0013
Therefore percent of days the factory produced 20320 or more was
0.13%
c)On what percent of the days last ear did the factory produce 20,160 or fewer.
therefore we have to find P(X < 20160) or P(z <( (20160- 20200)/40))
therefore we have to find P(z < -40/40 ) or P(z < -1)
this is a left tailed test. therefore from z table we P(z < -1) we get P(Z < -1) = 0.1587
Therefore percent of days the factory produced 20160 or fewer was 15.87%
