MEAN=50 STANDARD DEVIATION =7 P(3541)= *Round to 4 decimal as needed*
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Q: MEAN=50 STANDARD DEVIATION =7 P(3440)= *Round to 4 decimal as needed*
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P(35<x<63)=
P(x>41)=
*Round to 4 decimal as needed*
The mean is 50 and standard deviation is 7.
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- A set of 700 exam scores is normally distributed with a mean = 76 and standard deviation =Mean= 100 and standard deviation=10, what is probability of x<80 or x>110Central Limit Theorem: When the maximum SAT score was 2400, scores had a mean = 1518 and standard deviation = 325. Assuming the scores are normally distributed, determine the probability of a score exceeding 1600. If 81 SAT scores are randomly selected, find the probability that their mean score exceeds 1600.
- OreThe mean of a distribution is 15 and the standard deviation is 3. Use Chebyshev's Theorem and the formula 1 - 1 / (k ^ 2) to determine the minimum percentage of values that will fall between 9 and 21Some students paid a private tutor to help them improve their results on a certain mathematical test. These students had a mean change in score of + 18 points, with a standard deviation of 68 points. In a random sample of 100 students who pay a private tutor to help them improve their results, what is the likelihood that the change in the sample mean score is less than 10 points? Click the icon to view the table of normal curve areas. Normal Curve Areas P(x< 10) = (Round to three decimal places as needed.) 0. .00 .01 .02 .03 .04 .05 .06 .07 .08 .09 .0000 .0398 .0793 .1179 .1554 .1915 .0 .1 .0040 .0438 .0080 .0478 .0871 1255 .0160 .0557 .0948 .1331 .1700 2054 .0120 .0517 .0199 .0596 0987 1368 .1736 2088 .0239 .0636 1026 .1406 1772 2123 0279 .0675 1064 1443 1808 2157 .0319 0714 .1103 1480 1844 2190 0359 .0753 1141 1517 1879 2224 .2 .3 .0832 .1217 .1591 1950 .0910 1293 1664 1628 .5 .1985 2019 .2257 .2580 .2881 .3159 3413 2291 2611 .6 .7 .8 .9 1.0 2324 2642 2939 3212 3461 2357 2673 2967…
- A set of data is normally distributed with mean 51 and standard deviation 5. Determine the value of X corresponding to a z-score of -1.75. ** Answer to two decimal places **Numbers are normal dıstrıbuted and average = 60, standard deviation = 5.What should be the boundaries that determine 75% of the distribution in the center? (I can't use exel)
