Statistics Questions

Statistics Questions
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Institution
Statistics Questions
Chapter 5
Question 5.11 (Using Z Distribution Table)
Z=X-XS Z=125-10015
Z=1.667
PZ>1.67=0.9525
1 – 0.9525 = 0.0475
0.0475*100=4.75%
4.75% of the IQ scores are above Kristen’s 125.
Z=X-XS Z=82-10015
Z=-1.2
PZ<-1.2=0.1151
0.1151*100=11.51%
11.51% of the IQ scores are below 82.
Z=X-XS Z=91-10015 and 109-10015
Z=-0.6 and 0.6
P-0.6<Z=0.2743
P0.6>Z=0.7257
0.7257-0.2743= 0.4514
0.4514 *100=45.14%
45.15% of the IQ scores are within 9 points of the mean.
Z=X-XS Z=60-10015 and 140-10015
Z=-2.67 and 2.67
P-2.67<Z=0.0038
P2.67>Z=0.9962
0.9962 -0.0038= 0.9924
1- 0.9924=0.0076
0.0076 *100=0.76%
0.76% of the IQ scores are more than 40 points from the mean.
Question 5.13 (Using Z Distribution Table)
The probability closest to 0.98 is 2.06 which means the IQ score on the upper 2% is (0.98 * 15)+ 100 = 114.7
The probability closest to 0.1 is -2.32 which means the IQ score is (-2.32 * 15) + 100 = 134.8
The probability closest to 0.60 is 0.26 which means the IQ score on the upper 2% is (0.26 * 15) + 100 = 103.9
The probability closest to 0.025 is -2.81 and the probability closest to 0.975 is 2.81 which means the IQ score in the middle 95% is between (-2.81 * 15) + 100 = 57.85 and (2.81 * 15) + 100 = 142.15.
The probability closest to 0.01 is -2.33 and the probability closest to 0.99 is 2.33 which means the IQ score in the middle 99% is between -2.33*15+100=65.05 and 2.33*15+100=134.95Question 5.15 (Using Z Distribution Table)
The probability closest to 0.

05 is -1.64 which means the estimated number of hours for the shortest suffering 5% is -1.64*20+83=50.2
Z=X-XS Z=48-8320
Z=-1.75
PZ>-1.75=1-0.0401=0.9599
0.9599*100=95.99%
Z=X-XS Z=61-8320
Z=-1.1
PZ<-1.1=0.1357
0.1357*100=13.57%
The probability closest to 0.01 is -2.33 and the probability closest to 0.99 is 2.33 which means the estimated number of hours suffered by the extreme 1% on both ends of the mean are -2.33*20+83=36.4 hours and 2.33*20+83=129.6 hours.
Z=X-XS Z=24-8320 and 72-8320
Z=-2.95 and 0.55
P-2.95<Z=0.0016
P0.55>Z=0.7088
0.7088 -0.0016 = 0.7072
0.7072 *100=70.72%
The probability closest to 0.025 is -2.81 and the probability closest to 0.975 is 2.81 which means the estimated number of hours suffered by the middle 95% is between -2.81*20+83=26.8 hours and 2.81*20+83=139.2 hours.
Z=X-XS Z=48-8320 and 96-8320
Z=-1.75 and 0.65
P-1.75<Z=0.0401
P0.65>Z=0.7422
0.7422 -0.0401 = 0.7021
0.7021 *100=70.21%
The probability closest to 0.2 is -0.84 which means the medical researcher will work with those who suffered for more than -0.84*20+83=66.2 hours.
The probability closest to 0.03 is -1.88 and the probability closest to 0.97 is 1.88 which means the mild group consist of those who suffered fewer than -1.88*20+83=45.4 hours and the severe group consist of those who suffered more than 1.88*20+83=120.6 hours
Z=X-XS Z=61-8320
Z=1.1
PZ>-1.1=1-0.1357=0.8643
0.8643*100=86.43%
Z=X-XS Z=60-8320and 61-8320
Z=-1.15 and-1.1
PZ<-1.15=0.1251
PZ<-1.1=0.1357
0.1357- 0.1251=0.0106
0.0106*100=1.06%
Question 5.18 (Using Z Distribution Table)
Given that the distribution is positively skewed, the median is exceeded by the mean BMI score of 28.
Z=X-XS Z=25-284 and 29.9-284
Z=between-0.75 and 0.475
-0.75 < Z < 0.475 defines overweight
Z=X-XS Z=30-284
Z=0.5
Z > 0.5 defines obese
Chapter 8
Question 8.10
The sample used is not random since it is a convenience poll where people self-selected themselves to become participants.
To achieve random sampling, the sample has to have randomness whereby there is an equal chance of choosing any member of the population and external selection whereby participants are chosen instead of choosing to take the survey themselves. Consequently, the poll could be improved by calling randomized numbers rather than waiting for calls which helps achieve both randomness and external selection (Neuman, 2016).
Question 8.14
The probability of two boys = 13The probability of two girls = 13The probability of either two boys or two girls = 23Question 8.16
The probability of C = 15The probability of CI (in that order) for two guesses = 15*45=425=0.16The probability of CCC for three guesses = 15*15*15=1125=0.008The probability of III for three guesses = 45*45*45=16125=0.128Question 8.19
Excess radiation No excess radiation
Alarm 0.97 0.02
No alarm 0.01 0.03
The probability a sensor gives an incorrect report = 0.01+0.02=0.03The new probability the reactor shuts down due to simultaneous false alarms from both sensors = 0.97*0.97=0.941The new probability that excessive radiation is missed simultaneously by both sensors = 0.01*0.01=0.001Question 8.21
In a population of 1,000 with a 0.01 prevalence, actually 1,000*0.01=10 have the disease. Therefore, 1,000 –10=990 do not have breast cancer.
Has breast cancer Don’t have breast cancer (totals)
Test result positive 10*0.8=8990*0.1=99107
Test result negative 10*0.2=2 990*0.9=891893
(totals) 10 990 1000
The probability that a randomly selected woman will have a positive mammogram = 1071000=0.107The probability of having breast cancer, given a positive mammogram = 8107=0.075The probability of not having breast cancer, given a negative mammogram = 891893=0.998References
Neuman, W. L. (2016). Understanding research. Pearson.