Suppose x has a normal distribution with mean 48 and standar

Suppose x has a normal distribution with mean -48 and standard deviation 15. Describe the distribution of for sample size n-4. (Round or to two decimal places.) values Describe the distribution of x values for sample size n-16. (Round to two decimal places.) Describe the distribution of x values for sample size n 100. (Round o to two decimal places.) = ox = How do the x distributions compare for the various samples sizes? O The means are the same, but the standard deviations are increasing with increasing sample size. O The means are the same, but the standard deviations are decresing with increasing sample size. O The standard deviations are the same, but the means are decreasing with increasing sample size. O The means and standard deviations are the same regardiess of sample size. he standard deviations are the same, but the means are increasing with increasing sample size

Solution

1) mean =48 ,sd = 15 / sqrt(4) =7.5 2) mean= 48, sd = 15 / sqrt(16) = 3.75 3) mean= 48 , sd = 15/10 = 1.5....... so, means are same but s.d \'s are decreasing with increasing sample size!
2) p(x <= 60) = p( z<= (60-49) / 6.5 ) = p(z<= 1.6923) = 0.9547056.....................

p(x>50) =1-p(x<=50) = 1 - p(z<= 0.1538) =1 - 0.5611163 = 0.4388837
p(50<=x<=60) = p( x<=60) -p(x<=50) = 0.9547056 - 0.5611163 = 0.3935893

3) z for which 14% area lies to the right = z for 0.86 confidence = 1.080319
63% lies betwee -z to +z..right? then in left side there are (100-63)/2 % = 18.5% area and 18.5% area to the right...the in between area is which we are looking for....= z (0.185) or z( 0.815) =+ or - 0.8964734

4) mean=8.5 , s.d= 2.5

p(x<10) = p( (x-mean) / s.d) <(10-8.5) / 2.5 ) =p(z<= 0.6) =0.7257469
p( x>5) = 1- p(x<=5) = 1 - p( z < -1.4 ) = 1 - 0.08075666 = 0.9192433
p(8 <x<15) = p(x<15) - p(x<8) =p( z < 2.6 ) - p ( z <= -0.2 ) = 0.9953388 - 0.4207403 =  0.5745985