This paper concentrates on the primary theme of 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 participan in which you have to explain and evaluate its intricate aspects in detail. In addition to this, this paper has been reviewed and purchased by most of the students hence; it has been rated 4.8 points on the scale of 5 points. Besides, the price of this paper starts from £ 79. For more details and full access to the paper, please refer to the site.

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?

μ = 523, σ = 103, n = 60, z = (x-bar - μ)÷(σ÷√n)

z = (300 - 523)÷(103÷√60) = -16.7704 -16.77039391

P(x-bar < 300) = P(z < -16.7704) = 0

z = (450 - 523)÷(103÷√60) = -5.4899 -5.4899

P(x-bar > 450) = P(z > -5.4899) = 1.00

Formula is on page 282