P([latex]\displaystyle\overline{{x}}[/latex] < 2) = 0 The probability that the mean stress score is less than two is about zero.

normalcdf [latex]\displaystyle{({1},{2},{3},\frac{{1.15}}{\sqrt{{75}}})}={0}[/latex]

Remember that the smallest stress score is one.

Let k = the 90th percentile.

Find k, where P([latex]\displaystyle\overline{{x}}[/latex] < k) = 0.90.

k = 3.2

The 90th percentile for the mean of 75 scores is about 3.2. This tells us that 90% of all the means of 75 stress scores are at most 3.2, and that 10% are at least 3.2.

invNorm[latex]\left (0.90,3,\frac{1.15}{\sqrt {75}} \right ) = 3.2[/latex]

For problems 3 and 4, ΣX = the sum of the 75 stress scores. Then, [latex]\displaystyle\sum{X}{\sim}{N}{[{({75})}{({3})},{(\sqrt{{75}})}{({1.15})}]}[/latex]

The mean of the sum of 75 stress scores is (75)(3) = 225

The standard deviation of the sum of 75 stress scores is
[latex]\displaystyle{(\sqrt{{75}})}[/latex](1.15) = 9.96
P(Σx < 200) = 0

The probability that the total of 75 scores is less than 200 is about zero.

normalcdf (75,200,(75)(3),[latex]\displaystyle{(\sqrt{{75}})}[/latex](1.15)).
Remember, the smallest total of 75 stress scores is 75, because the smallest single score is one.

Let k= the 90th percentile.

Find k where P(Σx < k) = 0.90.

k = 237.8

The 90th percentile for the sum of 75 scores is about 237.8. This tells us that 90% of all the sums of 75 scores are no more than 237.8 and 10% are no less than 237.8.

invNorm(0.90,(75)(3),[latex]\displaystyle{(\sqrt{{75}})}[/latex](1.15)) = 237.8

Find: [latex]\displaystyle{P}{(\overline{{x}}{>}{20})}[/latex]

[latex]\displaystyle{P}{(\overline{{x}}{>}{20})}={0.79199}[/latex] using normalcdf [latex]\displaystyle{({20},{1}\text{E99},{22},\frac{{22}}{\sqrt{{80}}})}[/latex]

The probability is 0.7919 that the mean excess time used is more than 20 minutes, for a sample of 80 customers who exceed their contracted time allowance.

Remember, 1E99 = 10⁹⁹ and –1E99 = –10⁹⁹. Press the EE key for E. Or just use 10⁹⁹ instead of 1E99.

Find P(x > 20). Remember to use the exponential distribution for an individual: [latex]\displaystyle{X}\text{~}{E}{x}{p}{(\frac{{1}}{{22}})}[/latex].[latex]\displaystyle{P}{({x}{>}{20})}={e}^{{{(-{({\frac{1}{22}})}{({20})})}}}{\quad\text{or}\quad}{e}^{{{(-{0.04545}{({20})})}}}={0.4029}[/latex]

[latex]\displaystyle{P}{({x}{>}{20})}={0.4029}\text{ but } {P}{(\overline{{x}}{>}{20})}={0.7919}[/latex].

The probabilities are not equal because we use different distributions to calculate the probability for individuals and for means.

When asked to find the probability of an individual value, use the stated distribution of its random variable; do not use the CLT. Use the CLT with the normal distribution when you are being asked to find the probability for a mean.

Let k = the 95th percentile. Find k where [latex]\displaystyle{P}{(\overline{{x}}{<}{k})}={0.95}[/latex]
k = 26.0 using invNorm[latex]\left ( 0.95,22,\frac{22}{\sqrt {80}} \right )[/latex] = 26.0

The 95th percentile for the sample mean excess time used is about 26.0 minutes for random samples of 80 customers who exceed their contractual allowed time.

Ninety five percent of such samples would have means under 26 minutes; only five percent of such samples would have means above 26 minutes.

We have, [latex]\displaystyle{\mu}_{x}={\mu}=[/latex] 2 and [latex]{\sigma}_{x}=\frac{{\sigma}}{{\sqrt{n}}}=\frac{{0.5}}{{10}}=0.05[/latex]. Therefore:

1. 50th percentile = μ_x = μ = 2
2. 25th percentile = invNorm(0.25,2,0.05) = 1.97
3. 75th percentile = invNorm(0.75,2,0.05) = 2.03

We have [latex]\displaystyle{\mu}_{\sum{x}}=n({\mu}_{x})={100}({2})= 200[/latex] and [latex]{\sigma}_{\mu{x}}={\sqrt n}({\sigma}_{x})=10(0.5)=5[/latex]. Therefore:

1. 50th percentile = [latex]\displaystyle{\mu}_{\sum{x}}=n({\mu}_{x})[/latex] = 100(2) = 200
2. 25th percentile = invNorm(0.25,200,0.05) = 196.63
3. 75th percentile = invNorm(0.75,200,0.05) = 203.37

[latex]{P}{(1.75 < {\overline{x}} < {1.85})}[/latex] = normalcdf(1.75,1.85,2,0.05) = 0.0013

Using the z-score equation, [latex]\displaystyle{z}=\frac{{\overline{x}-{\mu}_{\overline{x}}}}{{{\sigma}_{\overline{x}}}}[/latex].

Solving for x, we have x = 2(0.05) + 2 = 2.1

The IQR is 75th percentile – 25th percentile = 203.37 – 196.63 = 6.74

P([latex]\overline{{x}}[/latex] < 35) = normalcdf(-E99,35,30.9,1.8) = 0.9886

P([latex]\overline{{x}}[/latex] > 50) = normalcdf(50, E99,30.9,1.8) ≈ 0.

For this sample group, it is almost impossible for the group’s average age to be more than 50. However, it is still possible for an individual in this group to have an age greater than 50.

P(Σx ≥ 1,600) = normalcdf(1600,E99,1514.10,63) = 0.0864

P(Σx ≤ 1,595) = normalcdf(-E99,1595,1514.10,63) = 0.9005.

This means that there is a 90% chance that the sum of the ages for the sample group n = 49 is at most 1,595.

The 95th percentile = invNorm(0.95,30.9,1.1) = 32.7.

This indicates that 95% of the sex workers in the sample of 65 are younger than 32.7 years, on average.

The 90th percentile = invNorm(0.90,2008.5,72.56) = 2101.5.

This indicates that 90% of the sex workers in the sample of 65 have a sum of ages less than 2,101.5 years.

Let X = the number that favor a charter school for grades K trough 5. X ~B(n, p) where n = 300 and p = 0.53. Since np > 5 and nq > 5, use the normal approximation to the binomial. The formulas for the mean and standard deviation are μ = np and [latex]\sigma = {\sqrt {npq}}[/latex]. The mean is 159 and the standard deviation is 8.6447. The random variable for the normal distribution is Y. Y ~ N(159, 8.6447).

1. You include 150 so P(X ≥ 150) has normal approximation P(Y ≥ 149.5) = 0.8641.

normalcdf(149.5,10^99,159,8.6447) = 0.8641

2. You include 160 so P(X ≤ 160) has normal approximation P(Y ≤ 160.5) = 0.5689. 

normalcdf(0,160.5,159,8.6447) = 0.5689

3. You exclude 155 so P(X > 155) has normal approximation P(y > 155.5) = 0.6572.

normalcdf(155.5,10^99,159,8.6447) = 0.6572

4. You exclude 147 so P(X < 147) has normal approximation P(Y < 146.5) = 0.0741.

normalcdf(0,146.5,159,8.6447) = 0.0741

5. P(X = 175) has normal approximation P(174.5 < Y < 175.5) = 0.0083.

normalcdf(174.5,175.5,159,8.6447) = 0.0083