Question

# 1.) A distribution of values is normal with a mean of 210 and a standard deviation...

1.)

A distribution of values is normal with a mean of 210 and a standard deviation of 3.

Find the interval containing the middle-most 78% of scores:

Enter your answer accurate to 1 decimal place using interval notation. Example: (2.1,5.6)

Hint: To work this out, 1) sketch the distribution, 2) shade the middle 78% of the data, 3) label unkown data values on the horizontal axis just below the upper and lower ends of the shaded region, 4) calculate the percentage (area) to the left of the lowermost unknown data value, 5) calculate the TOTAL percentage (area) to the LEFT of the uppermost unknown data value, 6) use invnorm to calculate the lowermost data value, 7) use invnorm to calculate the uppermost data value. The data values you get from steps 6 and 7 are used to answer the question.

2.)

Thinking it might be helpful, you ask the TCC Institutional Researcher for data. He doesn’t have the data you need, but he does tell you that the population mean GPA is 2.82 with population standard deviation of 1.07 for all students who have ever attended Tacoma Community College.

Find the z-score for a GPA of 3.24.
Round to 2 decimal places.

Find the probability a student at random will have a GPA over 3.24.
Round to 4 decimal places.

79.1% of students will have a GPA lower than:
Round to 2 decimal places.

If 5 students are chosen at random, what is the probability their mean GPA will be more than 3.24?
% Round to the nearest whole percent.

3.)

A population of values has a normal distribution with μ=172.6μ=172.6 and σ=25.9σ=25.9. You intend to draw a random sample of size n=53n=53.

Find the probability that a single randomly selected value is between 176.9 and 177.2.
P(176.9 < X < 177.2) =

Find the probability that a sample of size n=53n=53 is randomly selected with a mean between 176.9 and 177.2.
P(176.9 < < 177.2) =

2) Given that

mean μ = 2.82, standard deviation σ = 1.07 , x = 3.24

Z-score = (X-μ)/σ = (3.24-2.82)/1.07 = 0.39

P(X > 3.24) = P(z > 0.39) = 0.3482

=> X = μ-Z×σ = 2.82-(0.8099×1.07) = 1.95

3)

μ=172.6 σ=25.9 n=1

P(176.9 < X < 177.2) = P( (176.9-172.6)×(√1/25.9)) < z < P( (177.2-172.6)×(√1/25.9)

= P(0.166 < z < 0.177)

= P(z<0.177)-P(z<0.166)

= 0.5702 - 0.5659

= 0.0043

μ=172.6 σ=25.9 n=53

P(176.9 < X̅  < 177.2) = P( (176.9-172.6)×(√53/25.9)) < z < P( (177.2-172.6)×(√53/25.9)

= P(1.20 < z < 1.29)

= P(z<1.29)-P(z<1.20)

= 0.9014 - 0.8849

= 0.0165

μ=172  σ=25.9 n=53

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