Standard Normal Probabilities Table Table entry for z is the area under the stan

Question

Standard Normal Probabilities Table Table entry for z is the area under the standard normal curve to the let of z. 0001 02 03 04 05 06 07 08 .0 .5000 .5040 0.1 5398 5438 5080 5120 5160 .5199 5239 5279 .5319 5359 .5478 5517 5557 5596 5636 5675 5714 5753 0.2.5793 .5832 5871 5910 5948 5987 .6026 6064 .6103 6141 6179 6217 6255 6293 63316368 .6406 6443 .6480 6517 0.4.6554 .6591 .6628 .6664 .6700 .67366772 .6808 .6844 .6879 .6915 6950, ..698S .. .7019 7054 . .7088 .7123·157-,71901 ,72241 0.6 7257 .7291 7324 7357 7389 7422 .7454 .7486 7517 .7549 0.8 .7881 .7910 7939 7967 7995 8023 8051 .8078 81068133 0.98159 8186 821 8264 8289 8315 8340 LO8413 .8438 8461 .8485 8508 .8531 8554 .8577, .8599.8621 8790 88108830 98749 1.2 8849 .8869 8888 8907 .8925 8944 8962 8980 .8997 9015 9115, 9131m,9147 9162-1 ,91771 ·Q.] 1.4 .91 92 .9207.9222 .9236 .9251 .9265 .9279 .9292 .9306 .9319 1.6 .94529463 9474 9484 9495.9505 9515 9525 .9535 .9545 1.8 .9641 96499656 .9664 9671 9678 .9686 9693 9699 970 2.0 9772 .9778 9783 9788 9793 9798 .9803 9808 9812 .9817 2.2 .9861 9864 .9868 9871 9875 .9878 9881 .9884 9887 9890 2.4 .9918 .9920 .9922 .9925 .9927 .9929 .9931 9932 .9934 .9936 2.6 .9953 9955 .9956 .99579959 9960 9961 9962 9963 9964 591 9599 9608 961 9738 9744 97509756 9830 9834 .9838 298469850 9896 9898 9901 9904 9906 9909 9911 9913 9952 2.5 9938 9940 9941 9943 9945 9946 9948 9949 9951 9965 .9966 .9967 .9968 .9969 . .9970 .9971 , .9972 997399741 9974 .9975 9976 9977 9977 9978 9979 9979 .9980 9981 9981 .9982 9982 9983 9984 .99849985 9985 9986 9986 2.8 . 3.0 .9987 .9987 .9987 .988 .9988 9989 .9989 9989 9990 .9990 3.1 .9990 .9991 9991 9991 9992 9992 9992 9992 9993 9993 3.2 9993 .9993 9994 9994 9994 9994 9994 9995 .9995 .9995 3.3 .9995 .9995 .9995 .9996 9996 9996 9996 9996 .9996 9997 3.4 9997 .9997 9997 .9997 .9997 9997 .9997 .9997 9997 9998

Explanation / Answer

2)

1)

therefore probabiliy that student spent less than 15 Hours:

2)

probabiliy that student spent more than 22 Hours:

3)

4)

for 75th pecentile: z =0.67

therefore corresponding score =mean +z*std deviaiton =20+0.67*4.2=22.8

5)

for 20th pecentile: z =-0.84

therefore corresponding score =mean +z*std deviaiton =20-0.84*4.2=16.5

for normal distribution z score =(X-)/ here mean=       = 20 std deviation   == 4.200

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