An IQ test is designed so that the mean is 100 and the standard deviation is 1414 for the population of normal adults. Find the sample size necessary to estimate the mean IQ score of statistics students such that it can be said with 9090% confidence that the sample mean is within 66 IQ points of the true mean. Assume that sigmaσequals=1414 and determine the required sample size using technology. Then determine if this is a reasonable sample size for a real world calculation.

37

Step-by-step explanation:

The first thing is to calculate critical z factor

the alpha and the critical z score for a confidence level of 90% is calculated as follows:

two sided alpha = (100% - 90%) / 200 = 0.05

critical z factor for two sided alpha of .05 is calculated as follows:

critical z factor = z factor for (1 - .05) = z factor for (.95) which through the attached graph becomes:

critical z factor = 2.58

Now we have the following formula:

ME = z * (sd / sqrt (N) ^ (1/2))

where ME is the margin of error and is equal to 6, sd is the standard deviation which is 14 and the value of z is 2.58

N the sample size and we want to know it, replacing:

6 = 2.58 * (14 / (N) ^ (1/2))

solving for N we have:

N = (2.58 * 14/6) ^ 2

N = 36.24

Which means that the sample size was 37.

x + 107

step-by-step explanation:

let's say that the number he had before the mistake was x. then, he will have 107 dollars more than this, so the answer is x + 107.

the answer is 510

step-by-step explanation:

15 • 13 = 195

15 • 12 = 180

15 • 5 = 75

12 • 5 = 60

195 + 180 = 375

375 + 75 = 450

450 + 60 = 510