What is the GCF of 27 andd 63?
1 answer:
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Answer:
The probability that the mean test score is greater than 290
P(X⁻ > 290 ) = 0.0217
Step-by-step explanation:
<u><em>Step(i):</em></u>-
Mean of the Population (μ) = 281
Standard deviation of the Population = 34.4
Let 'X' be a random variable in Normal distribution
Given X = 290
<u><em>Step(ii):-</em></u>
<em> The probability that the mean test score is greater than 290</em>
P(X⁻ > 290 ) = P( Z > 2.027)
= 0.5 - A ( 2.027)
= 0.5 - 0.4783
= 0.0217
The probability that the mean test score is greater than 290
P(X⁻ > 290 ) = 0.0217