Question

1.An airline records flight delays in and out of Chicago O'Hare airport for a year. The average delay for these flights is 9.05 minutes with a standard deviation of 4.04 minutes. For a sample of 61 flights, 23% of flights will have an average delay less than how many minutes?

a.There is not enough information

b. 12.03

c. 6.07

d.9.43

e.8.67

2.As a result of the subprime mortgage crisis of 2008, approximately 11.81% of new mortgage originations went into foreclosure by 2009. Suppose in the Kalamazoo area there were 2,587 homes in which the mortgage originated during this period. What is the probability that less than 12.37% of the homes would be foreclosed on?

a.0.1887

b.>0.999

c.0.8113

d. 42.5921

e.0.5000

Answer 1

Solution:

Question 1)

Given:

Mean = $\mu =9.05$

Standard Deviation =

g4.04

Sample size= n= 61

Find sample mean $\bar{x}$ such that:

$P(\bar{X}<\bar{x}) = 0.23$

Thus find z value such that:

P( Z< z) =0.23

Look in z  table for Area = 0.2300 or its closest area and find corresponding z value.

,02 .00 .01 .03 .04 .05 .06 .07 .08 .09 Z .0фз .0004 .006 .db08 0012 0016 0023 0031 .0041 .0055 -3.4 |-3.3 -3.2 -3.1 -3.0 -2.9 -2.8 -2.7 -2.6 -2.5 -2,4 -2.3 -2.2 -2.1 -2.0 -1.9 -1.8 -1.7 -1.6 -1.5 -1.4 -1.3 -1,2 -1.1 -1.0 -0.9 -0.8 -0.7 ,0003 .0003 .0003 .0003 .0003 .0003 0002 .0003 .0003 0004 .0003 .0004 .0004 .0005 .0005 .0005 0006 0004 .0004 .0005 0007 .0007 .0006 0006 0006 .0005 .0005 .0008 0008 0009 .0013 0008 0007 .0007 .0010 .0009 0013 0009 0010 0013 0012 .0011 0011 0011 .0010 0018 .0025 .0017 .0016 .0015 .0015 .0014 .0014 .0019 0018 .0021 .0020 .0019 .0026 0024 .0023 .0022 .0021 .0030 .0040 .0035 .0034 .0033 .0032 .0029 .0028 .0027 .0026 .0039 .0038 0037 0036 0047 0045 .0060 .0044 0043 .0049 0066 0052 .0048 .0062 .0059 .0057 .0054 .0051 .0073 .0096 .0064 .0080 .0075 .0071 .0069 0068 .0082 0078 0089 .0087 .0084 .0107 .0104 .0102 .0099 0094 0091 0139 0136 0132 .0129 .0125 .0122 0119 .0116 .0113 .0110 0166 0158 0154 15 0146 143 .0179 .0174 0222 .0170 .0207 0262 0329 0409 0505 .0618 0149 .0901 1075 1271 1492 1736 2005 .0202 0197 .0250 0192 .0228 0217 .0212 0188 .0183 .0268 0287 0281 0256 .0244 .0239 .0233 .0274 .0301 0294 0307 0359 .0351 0344 0336 0322 .0314 .0446 .0436 .0427 0418 .0401 .0392 .0384 0375 .0367 .0516 0495 .0485 .0475 0465 .0455 0548 .0537 .0526 .0668 .0655 .0643 .0630 .0606 .0594 .0582 .0571 .0559 .0694 .0681 0823 .0808 .0793 .0764 .0735 0721 .0708 .0778 0853 .0838 .0968 1151 .0951 0934 .0918 .0885 0869 1131 1112 .1093 1292 1056 1038 1020 .1003 0985 1230 .1446 1685 .1190 1335 .1251 1210 1170 1357 1314 1587 .1562 1539 1515 1469 1423 1401 1379 1611 1841 .1814 .1762 .1711 1660 .1635 1788 1977 .2266 1922 1894 1867 2119 2090 .2061 2258 .2033 2927 1949 .2177 2420 2296 2236 2206 2148

Area -0.2296 is closest to 0.2300 and it corresponds to -0.7 and 0.04

thus z = -0.74

Now use following formula to find $\bar{x }$ value:

$\bar{x } = \mu + z \times \sigma /\sqrt{n}$

$\bar{x } = 9.05 + (-0.74) \times 4.04 /\sqrt{61}$

$\bar{x }= 9.05 + (-0.74) \times 4.04 /7.81025$

$\bar{x } = 9.05 - 0.38$

$\mathbf{{\color{DarkOrange} \bar{x }=8.67}}$

Question 2)

Given:

p = 0.1181

n = 2587

Find:

$P(\hat{p}<0.1237)=...........?$

Find z score for 0.1237

$z= \frac{\hat{p}-p}{\sqrt{\frac{p\times (1-p)}{n}}}$

$z= \frac{0.1237-0.1181}{\sqrt{\frac{0.1181 \times (1-0.1181)}{2587}}}$

$z= \frac{0.0056 }{\sqrt{\frac{0.1181 \times 0.8819 }{2587}}}$

$z= \frac{0.0056 }{\sqrt{ 0.00004026 }}$

$z= \frac{0.0056 }{0.0063451 }$

Thus we get:
$P(\hat{p}<0.1237)=P( Z < 0.88)$

Look in z table for z =0.8 and 0.08 and find corresponding area.

.00 .01 .02 .03 .04 .05 .06 .07 .08 .09 Z $319 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 5000 5040 5080 .5120 5160 5199 5239 .5279 5359 5517 5557 5714 .103 .480 344 7190 7517 823 .5478 .5596 5636 6026 5675 .5753 .5398 .5438 5987 6064 6443 5871 5910 5948 6141 5793 5832 6255 .6293 .6331 6368 6406 6517 .6179 6217 6628 6664 .6700 .6736 6772 .6808 6879 6554 6591 7054 7224 7019 7088 .7123 7157 .6915 6950 .6985 7357 7422 7486 7389 7454 7549 7257 7291 7324 7673 7704 7734 7764 7794 8078 7852 7580 7611 7642 7930 7967 7995 8023 8051 .8106 .8133 7881 7010

P( Z < 0.88) = 0.8106

closest answer is: 0.8113 thus correct answer is: 0.8113

Excel command gives accurate answer:

=NORM.S.DIST( z ,cumulative)

=NORM.S.DIST(0.882575,TRUE)

=0.8113