Question

2.) High school seniors' SAT scores are normally distributed with μ = 1050 and σ = 100. If a student is selected at random, find the probability that her SAT score is: 

a.) above 1200 b.) below 890 c.) between 1000 and 1100 

d.) What SAT score separates the smartest 4% of students? 

e). If 18 seniors are selected, find the probability that their mean SAT score is above 1150 

3.) A survey of 200 college students revealed that 160 of them eat dessert daily. Construct a 95% confidence interval for the population proportion of college students who eat dessert daily

Answer 1

Hi, We are supposed to answer one question at a time. Since you have not mentioned which question to answer. I am answering the first one. Please repost the remaining question that you would like to be answered.

2.

Given Information:

Mean = 1050

Standard Deviation = 100

To calculate the probabilities, the z score needs to be calculated first using following formula:

X - u z= σ where, X= Target Value z=z score o = standard deviation

a) The probability that SAT score is above 1200

Z = 1200-1050 =1.5 100

A(z) probability of a normal random variable not being more than a standard deviations above its mean. Values of z of particular importance: 1.645 1.960 2.326 2.576 3.090 3.291 A(3) 0.9500 0.9750 0.9900 0.9950 0.9990 0.9995 Lower limit of right 5% tail Lower limit of right 2.5% tail Lower limit of right 1% tail Lower limit of right 0.5% tail Lower limit of right 0.1% tail Lower limit of right 0.05% tail -4 -3 -2 -1 0 1 Z 2 3 4 0.00 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 1.1 1.2 0.5000 0.5398 0.5793 0.6179 0.6554 0.6915 0.7257 0.7580 0.7881 0.8159 0.8413 0.8643 0.8849 0.9032 0.9192 0.9332 0.9452 0.9554 0.9641 0.9713 0.9772 ni 0.5040 0.5438 0.5832 0.6217 0.6591 0.6950 0.7291 0.7611 0.7910 0.8186 0.8438 0.8665 0.8869 0.9049 0.9207 0.9345 0.9463 0.9564 0.9649 0.9719 0.9778 na 0.5080 0.5478 0.5871 0.6255 0.6628 0.6985 0.7324 0.7642 0.7939 0.8212 0.8461 0.8686 0.8888 0.9066 0.9222 0.9357 0.9474 0.9573 0.9656 0.9726 0.9783 0.5120 0.5517 0.5910 0.6293 0.6664 0.7019 0.7357 0.7673 0.7967 0.8238 0.8485 0.8708 0.8907 0.9082 0.9236 0.9370 0.9484 0.9582 0.9664 0.9732 0.9788 0001 0.5160 0.5557 0.5948 0.6331 0.6700 0.7054 0.7389 0.7704 0.7995 0.8264 0.8508 0.8729 0.8925 0.9099 0.9251 0.9382 0.9495 0.9591 0.9671 0.9738 0.9793 0.5199 0.5596 0.5987 0.6368 0.6736 0.7088 0.7422 0.7734 0.8023 0.8289 0.8531 0.8749 0.8944 0.9115 0.9265 0.9394 0.9505 0.9599 0.9678 0.9744 0.9798 on 0.5239 0.5636 0.6026 0.6406 0.6772 0.7123 0.7454 0.7764 0.8051 0.8315 0.8554 0.8770 0.8962 0.9131 0.9279 0.9406 0.9515 0.9608 0.9686 0.9750 0.9803 0.5279 0.5675 0.6064 0.6443 0.6808 0.7157 0.7486 0.7794 0.8078 0.8340 0.8577 0.8790 0.8980 0.9147 0.9292 0.9418 0.9525 0.9616 0.9693 0.9756 0.9808 onro 0.5319 0.5714 0.6103 0.6480 0.6844 0.7190 0.7517 0.7823 0.8106 0.8365 0.8599 0.8810 0.8997 0.9162 0.9306 0.9429 0.9535 0.9625 0.9699 0.9761 0.9812 on 0.5359 0.5753 0.6141 0.6517 0.6879 0.7224 0.7549 0.7852 0.8133 0.8389 0.8621 0.8830 0.9015 0.9177 0.9319 0.9441 0.9545 0.9633 0.9706 0.9767 0.9817 1.3 1.4 1.5 1.6 1.7 1.8 1.9 2.0 21 nonno

The probability can be calculated as 1 - given area = 1 - 0.9332 = 0.0668

b) probability below 890

890-1050 = -1.6 II 100

.00 01 102 05 .09 -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 .0003 .0005 10007 .0010 .0013 .0019 .0026 .0035 1.0047 .0062 .0082 1.0107 .0139 .0179 .0228 1.0287 .0359 1.046 .0548 .0668 1.0808 .0968 1,0003 .0003 1.0005 .0005 1.0007,0006 1.0009 1.0009 1.0013 .0013 .0018 .0018 1.0025 20024 1.0034 .0033 1.0045 .004 1.0060 .0059 .0080 1.0078 1.0104 .0102 .0136 .0132 20174 .0170 1.0222.0217 1.0281 .0274 1.0351 1.0344 1.0436 1.0427 1.0537 .0526 .0655 1.0643 1.0793 .0778 .0951 .0934 .03 104 .0003 10003 ,0004 1.0004 1,0006 1.0006 .0009 1.0008 1.0012 20012 1.0017 .0016 1.0023 .0023 .0032 .0031 1.0043 1.0041 .0057 1.0055 1.0075 .0073 1.0099 .0096 1.0129 .0125 20166 .0162 1.0212 1.0207 1.0268 1.0262 1.0336 .0329 .0418 .0409 .0516 10505 .0630 .0618 1.0764 .0749 1.0918 .0901 1.0003 .0004 1.0006 1.0008 1.0011 1.0016 1.0022 1.0030 .0040 .0054 20071 .0094 20122 .0158 .0202 1.0256 1.0322 .0401 .0495 1.0606 1.0735 1,0885 1.06 .07 1.08 .0003 10003 .0003 .0004 1.0004 10004 .0006 1.0005 1.0005 10008 1,0008 .0007 1.0011 1.0011 .0010 1.0015 .0015 .0014 1.0021 .0021 1.0020 .0029 .0028 1.0027 .0039 1.0038 .0037 .0052 .0051 .0049 .0069 .0068 1.0066 .0091 .0089 1.0087 .0119 .0116 1.0113 1.0154 .0150 .0146 .0197 .0192 .0188 1.0250 .0244 1.0239 1.0314 1.0307 1.0301 1.0392 .0384 1.0375 1.0485 1.0475 1.0465 1.0594 .0582 1.0571 1.0721 10708 1.0694 .0869 .0853 1.0838 10002 .0003 1,0005 .0007 .0010 .0014 .0019 .0026 .0036 .0048 .0064 .0084 .0110 .0143 .0183 1.0233 .0294 .0367 1.0455 .0559 1.0681 .0823

In this case, the probability will be 0.0548

c) between 1000 to 1100

Z= 1000-1050 100 = -0.5

Z= 1100 - 1050 100 = 0.5

A(z) probability of a normal random variable not being more than a standard deviations above its mean. Values of z of particular importance: 1.645 1.960 2.326 2.576 3.090 3.291 A(3) 0.9500 0.9750 0.9900 0.9950 0.9990 0.9995 Lower limit of right 5% tail Lower limit of right 2.5% tail Lower limit of right 1% tail Lower limit of right 0.5% tail Lower limit of right 0.1% tail Lower limit of right 0.05% tail 4 3 -2 -1 0 1 Z 2 0.00 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.5000 0.5398 0.5793 0.6179 0.6554 0.6915 0.7257 0.7580 0.7881 0.8159 0.8413 0.8643 0.8849 0.9032 0.9192 0.9332 0.5040 0.5438 0.5832 0.6217 0.6591 0.6950 0.7291 0.7611 0.7910 0.8186 0.8438 0.8665 0.8869 0.9049 0.9207 0.9345 0.5080 0.5478 0.5871 0.6255 0.6628 0.6985 0.7324 0.7642 0.7939 0.8212 0.8461 0.8686 0.8888 0.9066 0.9222 0.9357 0.5120 0.5517 0.5910 0.6293 0.6664 0.7019 0.7357 0.7673 0.7967 0.8238 0.8485 0.8708 0.8907 0.9082 0.9236 0.9370 0.5160 0.5557 0.5948 0.6331 0.6700 0.7054 0.7389 0.7704 0.7995 0.8264 0.8508 0.8729 0.8925 0.9099 0.9251 0.9382 0.5199 0.5596 0.5987 0.6368 0.6736 0.7088 0.7422 0.7734 0.8023 0.8289 0.8531 0.8749 0.8944 0.9115 0.9265 0.9394 0.5239 0.5636 0.6026 0.6406 0.6772 0.7123 0.7454 0.7764 0.8051 0.8315 0.8554 0.8770 0.8962 0.9131 0.9279 0.9406 0.5279 0.5675 0.6064 0.6443 0.6808 0.7157 0.7486 0.7794 0.8078 0.8340 0.8577 0.8790 0.8980 0.9147 0.9292 0.9418 0.5319 0.5714 0.6103 0.6480 0.6844 0.7190 0.7517 0.7823 0.8106 0.8365 0.8599 0.8810 0.8997 0.9162 0.9306 0.9429 0.5359 0.5753 0.6141 0.6517 0.6879 0.7224 0.7549 0.7852 0.8133 0.8389 0.8621 0.8830 0.9015 0.9177 0.9319 0.9441 0.9 1.0 1.2 1.3 1.4 1.5

-1.2 -1.1 -1.0 -0.9 -0.8 -0.7 -0.6 0.5 -0.4 -0.3 -0.2 0.1 -0.0 1151 1357 1587 1.1841 12119 1.2420 1.2743 1.3085 3446 1.3821 .4207 .4602 .5000 1131 1335 1.1562 .1814 .2090 1.2389 1.2709 1.3050 13409 3783 .4168 .4562 .4960 1112 1314 1539 1.1788 .2061 1.2358 1.2676 3015 1.3372 1.3745 .4129 .4522 1.4920 1095 1.1292 1.1515 .1762 .2033 1.2327 1.2643 .2981 3336 3707 .4090 .4483 14880 1075 11271 11492 .1736 .2005 1.2296 12611 1.2946 3300 1.3669 .4052 1.443 .4840 1056 .1251 1469 .1711 .1977 1.2266 12578 .2912 1.3264 3632 1.4013 14404 1.4801 1.1038 1230 1446 1.1685 1.1949 1.2236 1.2546 .2877 3228 13594 3974 .4364 4761 1020 .1210 1423 .1660 .1922 .2206 1.2514 .2843 3192 .3557 3936 1.4325 4721 1003 .1190 1401 .1635 .1894 .2177 12483 .2810 13156 13520 1.3897 1.4286 .4681 1.0985 1170 1379 1.1611 .1867 .2148 1.2451 .2776 3121 3483 3859 .4247 .4641

The probability can be calculated as:

p = 0.6915 - 0.3085 = 0.383

d) The SAT score value that separates the 4% of students

The z score at p - 0.04 is 1.75

X – 1050 1.75 = 100 X = 1050 +1.75*100 = 1225

Therefore, the SAT score is 1225

e) following is the formula for calculating z score when sample size is given

X – u Z= o Jn where, X= Target Value z=z score o = standard deviation = sample size

Z= = = 4.24 1150-1050 100 18

By referring to the standard normal table, the p-value at z = 4.24 is 0.0001