Respuesta :
Answer:
The probability that operator A gets the job even though both operators have equal ability is 0.0001.
Step-by-step explanation:
We are given that an experiment is designed to test whether operator A or operator B gets the job of operating a new machine. If the sample means for the 75 trials differ by more than 5 seconds, the operator with the smaller mean gets the job.
Suppose the standard deviations of times for both operators are assumed to be 2 seconds.
The z score probability distribution for the two-sample normal distribution is given by;
Z = [tex]\frac{(\bar X_1-\bar X_2)-(\mu_1-\mu_2)}{\sqrt{\frac{\sigma_1^{2} }{n_1}+\frac{\sigma_2^{2} }{n_2} } }[/tex] ~ N(0,1)
where, [tex]\bar X_1[/tex] = sample mean for operator A
[tex]\bar X_2[/tex] = sample mean for operator B
[tex]\sigma_1[/tex] = standard deviations of times for operator A = 2 seconds
[tex]\sigma_2[/tex] = standard deviations of times for operator B = 2 seconds
[tex]n_1=n_2[/tex] = sample of independent trials for both operators = 75
Now, the probability that operator A gets the job even though both operators have equal ability is given by = P([tex]\bar X_1 -\bar X_2[/tex] > 5 seconds)
P([tex]\bar X_1 -\bar X_2[/tex] > 5 sec) = P( [tex]\frac{(\bar X_1-\bar X_2)-(\mu_1-\mu_2)}{\sqrt{\frac{\sigma_1^{2} }{n_1}+\frac{\sigma_2^{2} }{n_2} } }[/tex] > [tex]\frac{5-0}{\sqrt{\frac{2^{2} }{75}+\frac{2^{2} }{75} } }[/tex] )
= P(Z > 15.31) = 1 - P(Z [tex]\leq[/tex] 15.31)
= 1 - 0.9999 = 0.0001
As in the z table, the highest critical value for x is given for x = 4.40 so we will take this value's probability area for x = 15.31.
