31, 6.2.9 Question Help Find the indicated IQ score. The graph to the right depicts IQ scores of adults, and those scores are normally distributed with a mean of 100 and a standard deviation of 15. Click to view page 1 of the table. Click to view page 2 of the table. X That's incorrect. X X Use technology or the standard normal distribution table to find the z score that corresponds to a cumulative area from the left of 0.9. Then convert the standard The indicated IQ score, x, is 80.8 (Round to one decimal place as needed.) normal random variable z to the normal random variable x using the formula below, where u is the mean and o is the standard deviation. X = H + (z . 5) OKQuestion 31, 6.2.9 Question Help i Standard Normal Table (Page 2) . X Find the indicated IQ score. The graph to the right depicts IQ scores of adults, and those score Click to view page 1 of the table. Click to view page 2 of the table. POSITIVE z Scores Standard Normal (z) Distribution: Cumulative Area from the LEFT X .0o 01 02 .03 .04 .05 .06 07 .08 .09 The indicated IQ score, x, is 80.8 (Round to one decimal place 0.0 5000 5040 5080 5120 .5160 .5199 5239 5279 5319 5359 0.1 5398 5438 5478 .5517 .5557 .5596 .5636 5675 5714 5753 0.2 .5793 5832 5871 5910 5948 5987 6026 6064 6103 6141 0.3 6179 5217 6255 6293 6331 6368 6406 6443 6480 6517 0.4 6554 6591 5628 6664 6700 6736 677 6808 6844 6879 0.5 6915 6950 6985 7019 7054 .7088 7123 7157 7190 7224 0.6 .7257 7291 7324 7357 .7389 742 ,7454 7486 ,7517 7549 0,7 .7580 7611 .7642 7673 7704 7734 7764 7794 7823 785 0.8 .7881 7910 7939 7967 7995 8023 8051 8078 8106 8133 0.9 8159 8186 8212 8238 8264 8289 8315 8340 8365 8389 1.0 .8413 8438 8461 8485 8508 .8531 .8554 8577 .8599 862 1.1 8643 8665 8686 8708 8729 8749 8770 8790 8810 8830 1.2 8849 8869 8888 8907 8925 8944 8962 8980 8997 9015 1.3 .9032 9049 9066 9082 9099 9115 9131 9147 .9162 9177 1.4 9192 9207 922 9236 9251 9265 9279 9292 9306 9319 1.5 9332 9345 9357 .9370 9382 9394 .9406 9418 9429 944 16 9452 2474 9484 9495 * 9505 9515 9525 9535 9545 Enter your answer in the answer box and then click Check Answer. Print Done ? All parts showing Clear All Check AnswerQuestion 37, 6.2.23-T Question Help A survey found that women's heights are normally distributed with mean 63.3 in. and standard deviation 3.3 in. The survey also found that men's heights are normally distributed with mean 69.2 in. and standard deviation 3.9 in. Most of the live characters employed at an amusement park have height requirements of a minimum of 57 in. and a maximum of 62 in. Complete parts (a) and (b) below. a. Find the percentage of men meeting the height requirement. What does the result suggest about the genders of the people who are employed as characters at the amusement park? The percentage of men who meet the height requirement is 31.87 %. x That's incorrect. X (Round to two decimal places as needed.) Note that this problem is equivalent to finding the area under the standard normal curve between the two z-scores that correspond to the minimum and maximum requirements. First sketch a normal curve, labeling the mean and any specific x-values. Then, shade the region representing the desired probability. For each relevant value x that is a boundary for the shaded region, use the formula below to convert that value to the equivalent z-score, where u is the mean and o is the standard deviation. Then use technology or a standard normal distribution table to find the area of the shaded region. Multiply by 100% to find the percentage. Note that with many technologies, it is not necessary to first convert the x-value to the equivalent z-score. X - H Z= OK Enter your answer in the answer box and then click Check Answer. ?Question 38, 6H224-T Question Helpv A survey tound that women's heights are normally distributed with mean 62.1 in and standard deviation 3 6 in The survey also found that men's heights are normally distributed With mean 69 5 in. and standard deviation 3 6 in. Consider an executive jet that seats SIX With a doorway height of 55 6 in Complete parts (a) through (c) below a. What perwntage of adult men can t through the door Without bending? The percentage of men who can t without bending is D 01 % (Round to two decimal places as needed) 1). Does the door design With a height of 55.6 in. appear to be adequate? Why didnt the engineers design a larger door? A. The door design is adequate, because although many men Will not be able to t without bending, most women Will be able to t Without bending Thus, a larger door is not needed B. The door design is adequate, because the majority of people Will be able to t without bending. Thus, a larger door is not needed. 3C. The door design is inadequate, but because the jet is relatively small and seats only six people, a much higher doorwould require major changes in the design and cost of the jet, making a larger height not practical D. The door design is inadequate, because every person needs to be able to get into the aircraft without bending. There is no reason why this should not be implemented. 1:. What doorway height would allow 40% of men to t without bending? The doorway height that would allow 40% of men to t Without bending is in (Round to one decimal place as needed ) Enteryour answer in the answer box and then click Check Answer. Question 39, 6.2.31 Question Help The lengths of pregnancies are normally distributed with a mean of 266 days and a standard deviation of 15 days. a. Find the probability of a pregnancy lasting 308 days or longer. b. If the length of pregnancy is in the lowest 2%, then the baby is premature. Find the length that separates premature babies from those who are not premature. Click to view page 1 of the table. Click to view page 2 of the table. a. The probability that a pregnancy will last 308 days or longer is .0026 . (Round to four decimal places as needed.) X That's incorrect. X b. Babies who are born on or before 235.25 days are considered premature. (Round to the nearest integer as needed.) Use the normal distribution table to find the z-score corresponding to the given area. Then use the fact that x = u + (z . ) to find the length associated with the given percentage. OK Enter your answer in the answer box and then click Check Answer. ? All parts showing Clear All Check AnswerQuestion 43, 6.4.17-T Question Help An engineer is going to redesign an ejection seat for an airplane. The seat was designed for pilots weighing between 150 lb and 191 lb. The new population of pilots has normally distributed weights with a mean of 159 lb and a standard deviation of 31.1 lb. Click here to view page 1 of the standard normal distribution. Click here to view page 2 of the standard normal distribution. a. If a pilot is randomly selected, find the probability that his weight is between 150 lb and 191 lb. X That's incorrect. X The probability is approximately .4626 . (Round to four decimal places as needed.) b. If 37 different pilots are randomly selected, find the probability that their mean weight is between 150 lb and 191 lb. In this case, the desired probability is for the mean of a sample. Therefore, use the The probability is approximately .8110 (Round to four decimal places as needed.) central limit theorem. OK Enter your answer in the answer box and then click Check Answer. ?Question 48, 6.5.19-T Question Help v a Asample of human brain volumes (cma) is given below Use the given data values to identify the corresponding 2 scores that are used for a normal duantile plot, then identify the coordinates of each point in the normal ouantile plot Construct the normal quantile plot, then determine whether the data appear to be from a population With a normal distribution 1087 1040 1067 1087 1025 952 1079 1418 l; List the z scores for the normal quantile plot -1.1 , -0.54 , -0.42 , -0.21 , -0.12 -0.06 , 2 50 (Round to two decimal places as needed. Use ascending order) . -0.06 x Try again. At least one of your answers is incorrect. 'Mth a sample of size n, each value 1 represents a portion of E of the sample. Using the known sample size n, find the 1 3 5 values of ' E' E' and so on, until reaching n values. Then use the standard normal distribution to nd the z scores corresponding to the cumulative left areas. These are the z scores that are expected from a normally distributed sample Remember to round to two decimal places and to use ascending order