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Normal Probability

See attached file and please solve calculations in red.

2 Using the descriptive statistics data determined during Week One's weekly problem
discussion, the mean for EI followed a standard distribution with a mean of 132.83 and a
standard deviation of 15.68. If we selected another random sample of 50 participants,

a What is the likelihood of selecting a sample with a mean EI score of at least 134?
b What is the likelihood of selecting a sample with a mean EI score of more than 128?
c What is the likelihood of selecting a sample with a mean EI score of more than 128 but less than 134?

Please show likelihood as a decimal with two decimal places.

μ = 132.83, σ = 15.68, n = 50, z = (x-bar - μ)÷(σ÷√n)

a z = (134 - 132.83)÷(15.68÷√50) = 0.5276 0.5276
P(x-bar ≥ 134) = P(z > 0.5276) = 0.30

b z = (128 - 132.83)÷(15.68÷√50) = -2.178 -2.178
P(x-bar > 128) = P(z > 0.5276) = 0.99

c P(128 < x-bar < 134) = P(-2.178 < z < 0.5276) = 0.69

Formula is on page 282

3 The mean Verbal SAT score for Division I student-athletes is 523 with a standard deviation of
103. If you select a random sample of 60 of these students, what is the probability the mean is
below 300? Above 450?

&#956; = 523, &#963; = 103, n = 60, z = (x-bar - &#956;)÷(&#963;÷&#8730;n)

z = (300 - 523)÷(103÷&#8730;60) = -16.7704 -16.77039391
P(x-bar < 300) = P(z < -16.7704) = 0

z = (450 - 523)÷(103÷&#8730;60) = -5.4899 -5.4899
P(x-bar > 450) = P(z > -5.4899) = 1.00

Formula is on page 282

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Solution Summary

The solution provides step by step method for the calculation of probability using the Z score. Formula for the calculation and Interpretations of the results are also included. Interactive excel sheet is included. The user can edit the inputs and obtain the complete results for a new set of data.

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