Question

0 Some students paid a private tutor to help them improve their results on a certain mathematical test. These students had a mean change in score of + 17 points, with a standard deviation of 66 points. In a random sample of 100 students who pay a private tutor to help them improve their results, what is the likelihood that the change in the sample mean score is less than 10 points? Click the icon to view the table of normal curve areas. P(x < 10) =(Round to three decimal places as needed.)

Answer 1

H = 17, 0 = 66, m = 100 We need to compute Pr(X < 10). The corresponding z-value needed to be computed: X-M z 10 - 17 66/1100 = -1.0606 oln Therefore, Pr(X < 10) = Pr (Z< ez (2 10 - 17 66/100 = Pr(Z < -1.0606) = 0.1444 The following is obtained graphically: Normal Distribution: Pr(Xbar<10)=0.1444 0.070 0.060 0.050 0.040 0.030 0.020 0.010 0.000 -10 un 5 10 15 20 25 30 35 40 45

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