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examination grades in an introductory statistics course is normally distributed, with a mean of 75 and a standard deviation o
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Solution:- Given that mean = 75 and standard deviation = 7

a. P(X < 87) = P((X-μ)/σ < (87-75)/7)
= P(Z < 1.7143)
= 0.9564

b. P(68 < X < 94) = P((68-75)/7 < (X-μ)/σ < (94-75)/7)
= P(-1 < Z < 2.7143)
= 0.8379

c. Z = 1.645

X = μ + Z*σ = 75 + 1.645*7 = 86.515 = 87

d.

=> Z = (89-75)/7 = 2

Z = (71-75)/2 = -0.57


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