3. In order to qualify for security officer’s training, recruits are tested for

Question

3. In order to qualify for security officer’s training, recruits are tested for stress tolerance. The scores are normally distributed, with a mean of 62 and a standard deviation of 8. If only the top 15% of recruits are selected, find the cut-off scores.

4. Waiting To be Seated. The average waiting time to be seated for dinner at a popular restaurant is 23.5 minutes, with a standard deviation of 3.6 minutes. Assume the variable is normally distributed. When the patron arrives at the restaurant for dinner, find the probability that the patron will have to wait the following times:

a) Between 15 and 22 minutes

b) Less than 18 minutes or more than 25 minutes

c) Is it likely that a person will be seated in less than 15 minutes? Justify your answer?

Explanation / Answer

3. From the information given, z=0.15, Xbar=62 and s=8. ubstituting the values in the following z score formula, obtain the Xi value that is the cut offf scores.

z=(Xi-Xbar)/s

0.15=(Xi-62)/8

Xi=63.2 (Answer).

4.

a. Find z scores corresponding to Xi=15 and Xi=22 using the preceding formula with Xbar=23.5 and s=3.6.

Z1=-2.36 and Z2=-0.42

Now using z table find the corresponding areas.0.4909 and 0.1628. Subtract the smaller one from larger one.

The required probability is 0.3281.

b. Z score corresponding to Xi=18 is -1.53 and z score crresponding to Xi=25 is 0.42.

Find areas corresponding to respective z scores. They are 0.4370 and 0.3372.

Required probability is 0.4370+0.3372=0.7742

c.Yes it i slikely that the person will be seated in less than 15 minutes as the probability is around 50%.

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