Question 6.10 A set of final examination marks in an introductory statistics unit is normally distributed with a mean of 73 and a standard deviation of 8.

What is the probability that a student obtains a mark between 65 and 89?

between 65 and 89 = 73 = 8 standardize x to z = (x - ) / P( 65 < x < 89) = P[( 65 - 73) / 8 < Z < ( 89 - 73) / 8] P( -1 < Z < 2) = 0.3413+0.4772 =0.8185 here, z2 = .9772 0. 5 = .4772

this is the value of z at 2

so, the result is 0.8185

Question 7.5The number of passengers passing through a large South East Asian airport in normally distributed with a mean of 110,000 persons per day and a standard deviation of 20,200 persons. If you select a random sample of 16 daysQuestion: What is the sampling distributions of the mean?

Answer:Normal Distribution and the mean, = 110,000 What is the probability that the sample mean is less than 98,000 passengers?

Answer:z = (x - ) / find the value of Zx = 110000 = 98000 = 20,200so, Z = 0.594now look at the Z table0.594 has the probability of 0.06so, the probability is 6%

What is the probability that the sample mean is between 102,000 and 104,500 passengers?

between 102000 and 104500 = 110000 = 20200 standardize x to z = (x - ) / P( 102000 < x < 104500) = P[( 102000 - 110000) / 20200 < Z < ( 104500 - 110000) /20200] P( -0.39604 < Z < -0.27228) = 0.1157+ (-0.1064) =0.0093 here, z-0.27 = 0.3936 0. 5 = -0.1064

this is the value of z at -0.27

so, the result is 0.0093 The probability is 60% that the sample mean will be between what two values symmetrically distributed around the population mean?

From Z table, for an area P = 0.60 symmetrically distributed around the mean , we need to know the Z-value for P = 0.60/2 = 0.30. Each tail occupies 20% of the area under the curve. For P(Z)