document.write( "Question 1151047: Suppose the number of students in a class for the Business Statistics program at a University has a mean of 23 with a standard deviation of 4.3. If 15 classes are selected randomly, find the probability that the mean number of students is between 20 and 30. \n" ); document.write( "
Algebra.Com's Answer #772669 by jim_thompson5910(35256)![]() ![]() ![]() You can put this solution on YOUR website! \n" ); document.write( " \n" ); document.write( " \n" ); document.write( "n = sample size \n" ); document.write( " \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "In this case, \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( "We're given two xbar values which are 20 and 30\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Compute the z score for the raw score xbar = 20 \n" ); document.write( " \n" ); document.write( " \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Repeat for xbar = 30 \n" ); document.write( " \n" ); document.write( " \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "We have \n" ); document.write( " \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Use a Z table such as this one \n" ); document.write( "http://www.z-table.com/ \n" ); document.write( "to find that, \n" ); document.write( "P(Z < -2.70) = 0.0035 \n" ); document.write( "P(Z < 6.30) = 1.00 \n" ); document.write( "note: if k is larger than 3.4, then P(Z < k) will be very very close to 1.00; this is especially true of z = 6.30 as its very distant from the center z = 0. So that is how I got P(Z < 6.30) = 1.00\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Subtract the values to find the area between the z scores \n" ); document.write( "P(A < Z < B) = P(Z < B) - P(Z < A) \n" ); document.write( "P(-2.70 < Z < 6.30) = P(Z < 6.30) - P(Z < -2.70) \n" ); document.write( "P(-2.70 < Z < 6.30) = 1 - 0.0035 \n" ); document.write( "P(-2.70 < Z < 6.30) = 0.9965\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Answer: 0.9965 \n" ); document.write( "This answer is approximate.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "side note: \n" ); document.write( "You can use a calculator like this one \n" ); document.write( "http://davidmlane.com/hyperstat/z_table.html \n" ); document.write( "to compute the area under the Z curve. Leave the mean and standard deviation as 0 and 1 respectively. Click the \"between\" radio button and type in the values -2.70 and 6.30 into the boxes. Then hit \"recalculate\" to have the answer come up. This can also be done on TI83 and TI84 calculators as well using the normalcdf function. \n" ); document.write( " \n" ); document.write( " |