From each of two normal populations with identical means and with standard deviations of 6.40 and 7.20, independent random samples of 64 observations are drawn. Find the probability that the difference between the means of the samples exceeds 0.6 in absolute value. math. stackexchange

0.31207

Step-by-step explanation: from the question, the distribution follows a normal distribution and population mean of both data set are equal.

I will also assume a 5% level of significance since any level of significance was not given in the question.

Population standard deviation of the first population (σ1) = 6.40

Population variance of the first population (σ1)² = 40.96

Population standard deviation of the second population (σ2) = 7.20

Population variance of the second population (σ2)² = 51.84

Sample size of first population (n1) = 64

Sample size of second population (n2) = 64

Difference between sample mean (x1 - x2) = 0.6.

Since population standard deviation is known, z test is used in getting the z score and a normal distribution table is used in getting the probability attached to the z score.

Z = (x1 - x2) /√{(σ1)²/n1 + (σ2)²/n2}

By substituting parameters, we have

Z = 0.6/√(40.96/64) + (51.84/64)

Z = 0.6/ √0.64 + 0.81

Z = 0.6/√1.45

Z = 0.6/1.2041

Z = 0.49

Since our difference in sample mean is greater than zero, it means that the probability of the z score (z=0.49) must be that of the one greater than 0.49 which is area towards the right of the distribution.

Do note that the probability value under z score to the left + probability value under z score to the right = 1

The table I'm using here is only giving probability of z scores toward the left (z<0.49)

Hence z>0.49 = 1 - z<0.49

From the table, z<0.49 = 0.68793

z>0.49 = 1 - 0.68793

z>0.49 = 0.31207

- 70

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