In a certain neighbourhood it is known that 12% of school learners are unemploye
ID: 3272350 • Letter: I
Question
In a certain neighbourhood it is known that 12% of school learners are unemployed. If a random sample of 150 school leavers is selected, what is the probability that the sample contains less than 15% unemployed? 1. 0.8708 2. 0.1500 3. 0 1292 4. 1.00 5. 0.95 It is important for airline to know the appropriate total weight of the baggage carried on each airplane. An airline researcher believes that the mean baggage weight for each adult is 60kg. To test his belief, he draws a random sample of 50 adult passengers and weights their baggage. He funds the sample mean to be 57.1 kg. If he knows that the population standard deviation is 10 kg, the probability that the average baggage weight is more than 57.1 is 1. 0.9798 2. 00202 3. 09772 4. 00979 5. -2.0506 Of the 100 people who were given a vaccine, 80 developed immunity to a disease. The 95% confidence interval of the proportion of people developing immunity is given by: 1. (0.7216, 0.8784) 2. (0.7342, 0.8658) 3. (0.7984, 0.8924) 4. 0.7480, 0.8106) 5. (0.7760, 0.8294)Explanation / Answer
Solution2:
p^=12%=0.12
n=150
P(p<0.15)
convert to z
0.15-0.12/sqrt(0.12(1-0.12)/150)
P(z<1.13)
=0.8708
ANSWER 0.8708
Solution3:
mean=60
n=sample size=50
sample mean=xbar=57.1
std dev=10
P(X>57.1)
convert to z
z=x-mean/sd(sqrt(n)
z=60-57.1/10/sqrt(50)
=2.9/10/sqrt(50)
=2.0506
P(z>2.0506)
=1-P(Z<2.0506)
=1-0.9798
=0.0202
ANSWER 0.0202
Solution4:
p^=sample prop=x/n=80/100=0.8
z for 95%=1.96
95% confidence interval for the population proportion of people developing immunity is given by
p^-zcritsqrt(p^(1-p^)/n),p^+zcritsqrt(p^(1-p^)/n)
=0.8-1.96sqrt(0.8(1-0.8)/100),0.8+1.96sqrt(0.8(1-0.8)/100)
=0.7216,0.8784
LOWER LIMIT=0.7216
UPPER LIMIT=0.8784
OPTION1