Question

An experiment is designed to test whether operator A or operator B gets the job of operating a new machine. Each operator is timed on 75 independent trials involving the performance of a certain task on the machine. If the sample means for the 75 trials differ by more than 5 seconds, the operator with the smaller mean gets the job. Otherwise, the experiment is considered to end in a tie. Suppose the standard deviations of times for both operators are assumed to be 2 seconds. What is the probability that operator A gets the job even though both operators have equal ability

Answer 1

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  =  $\frac{(\bar X_1-\bar X_2)-(\mu_1-\mu_2)}{\sqrt{\frac{\sigma_1^{2} }{n_1}+\frac{\sigma_2^{2} }{n_2} } }$  ~ N(0,1)

where, $\bar X_1$ = sample mean for operator A

$\bar X_2$ = sample mean for operator B

$\sigma_1$ = standard deviations of times for operator A = 2 seconds

$\sigma_2$ = standard deviations of times for operator B = 2 seconds

$n_1=n_2$ = 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($\bar X_1 -\bar X_2$ > 5 seconds)

    P($\bar X_1 -\bar X_2$ > 5 sec) = P( $\frac{(\bar X_1-\bar X_2)-(\mu_1-\mu_2)}{\sqrt{\frac{\sigma_1^{2} }{n_1}+\frac{\sigma_2^{2} }{n_2} } }$ > $\frac{5-0}{\sqrt{\frac{2^{2} }{75}+\frac{2^{2} }{75} } }$ )

                                    = P(Z > 15.31) = 1 - P(Z $\leq$ 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.