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Question 4 Unanswered On a recent English test, the scores were normally distributed with a mean of 74 and a standard deviation of 7. What proportion of the class would be expected to score between 60 and 80 points? A 0.7816 B 0.1729 C 0.8043 D 0.0228
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Answer #1

Mean = 74

S.D. = 7

P(60<x<80) to find

Z score at x = 60

Z = (X - μ) / σ
Z = (60 - 74) / 7
Z = -2

Z score at x = 80

Z = (X - μ) / σ
Z = (80 - 74) / 7
Z = 0.85714

P(60<x<80)

= P( -2 < X < 0.85714)

=P(X < 0.85714) - P(X< -2)

= 0.85714 - 0.0228

= 0.7816

P(60<x<80) = 0.7816

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