Question
Let X be normally distributed with mean = 3.3 and standard deviation = 2.8. [You may find it useful to reference the z table.]
a. Find P(X > 6.5). (Round "z" value to 2 decimal places and final answer to 4 decimal places.)
b. Find P(5.5 X 7.5). (Round "z" value to 2 decimal places and final answer to 4 decimal places.)
c. Find x such that P(X > x) = 0.0485. (Round "z" value and final answer to 3 decimal places.)
d. Find x such that P(x X 3.3) = 0.3531. (Negative value should be indicated by a minus sign. Round "z" value and final answer to 3 decimal places.)
TABLE 1 Standard Normal Curve Area Entries in this table provide cumulative probabilities, that is, the area der the curve to the left of -. For example, P(Zs 0.0000 0.0000 0.00000.0000 0.00000.0000 0.0000 0.0000 0.0000 0.0000 0.00010.0001 0.0001 0.00010.0001 0.0001 0.00010.0001 0.0001 0.00010.0001 0.0001 0.00010.0001 0.0001 0.0001 0.00030.0003 0.0003 0.00030.0003 0.0003 000030.0003 0.0003 0.0002 0.0005 0.0005 0.00050.0004 0.00040.0004 0.00040.0004 0.00040.0003 0.00070.0007 0.00060.0006 0.00060.0006 0.0006 0.0005 0.0005 0.0005 0.0010 0.0009 0.00090.0009 0.00080.0008 0.0008 0.0008 0.0007 0.0007 0.0010 0.0010 0.0019 0.0018 0.0018 0.001 0.0026 0.002 0.0020 0.0019 0.0034 0.003 0.0029 0.0028 0.002 0.0039 0.0038 0003 0.0069 0.0068 0.00660.0064 0.01190.0116 0.0113 0.0110 0.0060 0.0059 0.005 0.0078 0.007 0.0055 0.0054 0.005 0.0049 0.0048 0.0099 0.0096 0.0094 0.0089 0.0087 0.0170 0.0166 0.01 0.0150 0.0146 0.01 0.0314 | 0.0307 | 0.0301 | 0.0294 0.04460.0436 0.0427 0.04180.0409 0.0401 0.03920.0384 0.0375 0.0367 0.0548 0.0537 0.05260.0516 0.05050.0495 0.0485 0.0475 0.0465 0.0455 0.06680.0655 0.0643 0.06300.0618 0.0606 0.05940.0582 0.05710.0559 0.0808 0.0793 0.0708 00694 0.0918 0.09 0.11120.1093 0.1075 0.10560.1038 0.1020 0.10030.0985 0.13570.1335 0.1314 0.12920.1271 0.1251 0.12300.1210 0.1190 0.1170 0.15870.1562 0.15390.1515 0.1492 0.1469 0.1446 0.1423 0.1401 0.1379 0.18410.1814 0.1788 0.176 0.21190.2090 0.2061 0.203 0.1894 0.1867 0.2578 0.2546 0. 0.34460.3409 0.3372 0.33360.3300 0.3264 0.32280.3192 0.3156 0.3121 0.4168 | 0. 2910 OO.40 0.3974 0.3936 0.3897 0.3859 0.5000 0.4960 . 2010.4880| 0.484 0.4 0 0.50000.5040 0.5080 0.5120 0.5160 0.5199 0.5239 0.5279 0.5319 0.5359 0.5398 0.5438 0.54780.5 0 0.55960.5636 0.5675 0.5 0.6368 0.6406 0 0.7088 0.71 0.59100.5948 0.5987 0.60260.6064 0.6103 0.61 0.65910.6628 0.6664 0.6700 0.6736 0.6772 0.68080.684 0 0.8078 0.8106 0.81 0.81590.8186 0.8212 0.82380.82 0.8289 0.8315 0.83 0.8686 0.8708 0.87290.8749 0.8770 0.8790 0.8810 0.8830 0.8849 0.8869 0.8888 0.8907 0.8962 0.8980 0.89970.9015 0.9066 0.908 0.9099 0.9115 0.91 0.9370 0.9382 0.93940.9406 0.94180.9429 0.94 0 0.9599 0.9608 0.9616 0. 0.9678 0.9686 0.9693 0.9656 0.9664 0.9699 0.9706 0.97130.9719 0.9726 0.97 0.9750 0.9756 0.9761 0.9808 0.981 0.9868 0.987 0.98750.9878 0.9881 0.98840.9887 0.9890 0.9896 0.9898 0.9901 0.9906 0.9909 0.991 0.9918 0.9920 0.992 0.9938 0.9940 0.994 0.9946 0.9948 0.9949 0.995 0.9956 0.995 0.9959 0.9960 0.9961 0.9966 0.9967 0.9975 0.9976 0.99 0.9982 0.9968 0.9969 0.99700.99 0.99730.997 0.9978 0.9979 0.99790.9980 0.9981 0.9986 0.9986 0.9988 0.9988 0.9989 0.99890.9989 0.9990 0.9990 0.9982 0.9983 0.9984 0.99900.9991 0.999 0.9992 0.9992 0.999 0.9993 0.9995 0.9995 0.9995 0.9995 0.99950.9996 0.9996 0.9996 0.9996 0.9996 0.9996 0.9997 0.9998 0.9998 0.9998 0.99980.9998 0.9998 0.9998 0.9998 0.9998 0.9998 0.9998 . 99 | 0.9999 | 0.9999 | 0.9999 | 0.9999 | 0.9999 | 0.9999 0.9999 0.9999 0.9999 0.9999 0.9999 0.9999 0.9999 0.9999 0.9999 0.9999 0.9999 0.9999 0.9999 0.99990.9999 0.9997 0.9997 0.9997 0.9997 0.9998 0.9998 0.999 0.9999 0.9999 0.99 0.9999 0.9999
Explanation / Answer
for normal distribution z score =(X-mean)/std deviation
a)
P(X>6.5)=P(Z>(6.5-3.3)/2.8)=P(Z>1.14)=0.1271
b)
P(5.5<X<7.5)=P(0.79<Z<1.5)=0.9332-0.7852=0.1480
c)
for abvoe z score =1.66
therefore corresponding score =mean+z*std deviaiton=3.3+1.66*2.8=7.948
d)
for abvoe P(x<X<3.3)=0.3531
P(X<3.3)-P(X<x)=0.3531
P(Z<0)-P(X<x)=0.3531
0.5-P(X<x)=0.3531
P(X<x)=0.5-0.3531=0.1469
for abvoe critical z =-1.05
therefore x =mean+z*Std deviaiton=3.3-1.05*2.8= 0.36
