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© BrainMass Inc. brainmass.com October 25, 2018, 4:21 am ad1c9bdddf
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.
Statistics for Managers: Normal distribution, random sample, standard deviation
Given a normal distribution with a population mean of 100, a standard deviation of 10 and a sample size of n = 25, what is the probability that X is
a. less than 95?
b. between 95 and 97.5?
c. above 102.2?
d. There is a 65% chance that X is above what value?
Time spent using e-mail per session is normally distributed, with a population mean of 8 minutes and standard error of the mean is 2 minutes. If you select a random sample of 25 sessions,
a. what is the probability that the sample mean is between 7.8 and 8.2 minutes?
b. what is the probability that the sample mean is between 7.5 and 8 minutes?
c. If you select a random sample of 100 sessions, what is the probability that the sample mean is between 7.8 and 8.2 minutes?
d. Explain the difference in the results of (a) and (c).
You plan to conduct a marketing experiment in which students are to taste one of two different brands of soft drink. Their task is to correctly identify the brand tasted. You select a random sample of 200 students and assume that the students have no ability to distinguish between the two brands. (Hint: If an individual has no ability to distinguish between the two soft drinks, then the two brands are equally likely to be selected.)
a. what is the probability that the sample will have between 50% and 60% of the identifications correct?
b. the probability is 90% that the sample percentage is contained within what symmetrical limits of the population percentage?
c. what is the probability that the sample percentage of correct identifications is greater than 65%?
d. which is more likely to occur more than 60% correct identifications in the sample of 200 or more than 55% correct identifications in a sample of 1,000? Explain.
The fill amount of bottles of a soft drink is normally distributed, with a mean of 2.0 liters and a standard deviation of 0.05 liter. If you select a random sample of 25 bottles, what is the probability that the sample mean will be
a. between 1.99 and 2.0 liters?
b. below 1.98 liters?
c. greater than 2.01 liters?
d. the probability is 99% that the sample mean amount of soft drink will be at least how much?