Question

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 < *x¯* < 177.2) =

Enter your answers as numbers accurate to 4 decimal places.

Answer #1

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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Enter your answer accurate to 1 decimal place using interval
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Hint: To work this out, 1) sketch the distribution, 2) shade the
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Hint: To work this out, 1) sketch the distribution, 2) shade the
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