I will provide you with detailed feedback and you will be able to resubmit the lab assessment if you receive an incomplete.Revision Due Date:Monday, 12/5 at 8 pmThis lab assessment closes on...

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Answered Same DayNov 30, 2022

Answer To: I will provide you with detailed feedback and you will be able to resubmit the lab assessment if you...

Rajeswari answered on Nov 30 2022
65 Votes
114998 Assignment
Q.no.1
1. Spread is more for section 1 as z scores are high both negative and positive. Spread less means all scores are near mean. Hence group 2 is consistent here as all scores are sp
read around mean. Mean is here the central tendency as normal distribution is used.
2. First group having 77.5 mean more hence we can predict by central limit theorem that first group has more chance of having higher average than the second group.
3. Z scores are given in the list. No score is above 2 or below -2 in both the groups. Hence answer is no.
Q.no.2
1. 8 person z score is -1.59 while that of 23rd person is -0.78. Though actual score is high for 23rd person, comparing Z score we find that he has less risk of anxiety than 8th person.
2. P(SLE>40) = P(Z>0.932) = 0.5-0.3238 = 0.1762
This is not unusual as probability is significant as 17.62%
P(MSQ>125) = P(Z>2.029) =0.5-0.4788=0.0212
This is unusual as prob is less as 2%
3. P(SLE<21)=P(Z<-2.76) = 0.5-0.4971=0.0029
This is almost negligible hence this chance is unusual.
P(MSQ<110)=P(Z<-1.48)=0.5-0.4306=0.0694
This is not very unusual as prob is better as 6.94%
Q.no.3
A company hired a psychologist to assist their employees in their personal problems. The psychologist kept 1 file for each person they helped. Eight (8) employees sought out help for drug-related problems. Fifteen (15) employee needed help for family crisis problems. And twenty-two (22) employees needed help for miscellaneous reasons. The numbers are summarized below.
1. If one of the files is selected at random, what is the probability that it would involve a drug-related case? Interpret this probability.
Ans: 8/(8+15+22) = 8/45. i.e. there is 8/45 chance is there that a random person to be drug related case. Favourable events are 8 here and total 45 hence 8/45 is the probability.
2.
If one of the files is selected at random, what is the probability that it would involve a miscellaneous case? Interpret this probability.
Ans: 33/(8+15+22) = 33/45. i.e. there is 22/45 chance is there...
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