Math / Data & Statistics

QuestionAt a financial services company, the mean percent allocation for bonds in a retirement portfolio is 26.9 percent with a population standard deviation of 3.6 percent. A random sample of 35 retirement plan participants was taken and the probability that the mean bond percent for the sample will be at least 28 percent was determined.
Was the probability a Left-tail, Right -tail or Interval Probability? What is the probability that the mean bond percent for the sample will be at least 28 percent? Select all that apply below.
Select the two correct answers that apply below.
You may use a calculator or the portion of the $z$ -table given below. Round your answer try three decimal places if necessary. \begin{tabular}{|l|l|l|l|l|l|l|l|l|l|l|} \hline $z$ & 0.00 & 0.01 & 0.02 & 0.03 & 0.04 & 0.05 & 0.06 & 0.07 & 0.08 & 0.09 \\ \hline 1.6 & 0.945 & 0.946 & 0.947 & 0.948 & 0.949 & 0.951 & 0.952 & 0.953 & 0.954 & 0.954 \\ \hline 1.7 & 0.955 & 0.956 & 0.957 & 0.958 & 0.959 & 0.960 & 0.961 & 0.962 & 0.962 & 0.963 \\ \hline 1.8 & 0.964 & 0.965 & 0.966 & 0.966 & 0.967 & 0.968 & 0.969 & 0.969 & 0.970 & 0.971 \\ \hline 1.9 & 0.971 & 0.972 & 0.973 & 0.973 & 0.974 & 0.974 & 0.975 & 0.976 & 0.976 & 0.977 \\ \hline 2.0 & 0.977 & 0.978 & 0.978 & 0.979 & 0.979 & 0.980 & 0.980 & 0.981 & 0.981 & 0.982 \\ \hline 2.1 & 0.982 & 0.983 & 0.983 & 0.983 & 0.984 & 0.984 & 0.985 & 0.985 & 0.985 & 0.986 \\ \hline 2.2 & 0.986 & 0.986 & 0.987 & 0.987 & 0.987 & 0.988 & 0.988 & 0.988 & 0.989 & 0.989 \\ \hline \end{tabular}
Select all that apply: Left- Tail Right-Tail Interval $P(\bar{x}>28)=0.035$

Studdy Solution

STEP 1

1. The population mean percent allocation for bonds is $\mu = 26.9\%$ .
2. The population standard deviation is $\sigma = 3.6\%$ .
3. The sample size is $n = 35$ .
4. We are interested in the probability that the sample mean is at least $28\%$ .

STEP 2

1. Calculate the standard error of the mean.
2. Calculate the z-score for the sample mean of $28\%$ .
3. Determine the probability using the z-table.
4. Identify the type of probability (Left-tail, Right-tail, or Interval).

STEP 3

Calculate the standard error of the mean ( $SE$ ):
$SE = \frac{\sigma}{\sqrt{n}} = \frac{3.6}{\sqrt{35}}$
$SE \approx \frac{3.6}{5.916} \approx 0.608$

STEP 4

Calculate the z-score for the sample mean of $28\%$ :
$z = \frac{\bar{x} - \mu}{SE} = \frac{28 - 26.9}{0.608}$
$z \approx \frac{1.1}{0.608} \approx 1.809$

STEP 5

Using the z-table, find the probability that $\bar{x} > 28$ .
For $z = 1.81$ , the cumulative probability from the table is approximately $0.964$ .
Since we want $P(\bar{x} > 28)$ , we calculate:
$P(\bar{x} > 28) = 1 - P(\bar{x} \leq 28) = 1 - 0.964 = 0.036$

STEP 6

Identify the type of probability:
Since we are looking for the probability that the sample mean is greater than a certain value, this is a Right-tail probability.
The correct answers are: - Right-Tail - $P(\bar{x}>28)=0.036$

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Studdy Solution

STEP 1

1. The population mean percent allocation for bonds is $\mu = 26.9\%$ .
2. The population standard deviation is $\sigma = 3.6\%$ .
3. The sample size is $n = 35$ .
4. We are interested in the probability that the sample mean is at least $28\%$ .

STEP 2

1. Calculate the standard error of the mean.
2. Calculate the z-score for the sample mean of $28\%$ .
3. Determine the probability using the z-table.
4. Identify the type of probability (Left-tail, Right-tail, or Interval).

STEP 3

Calculate the standard error of the mean ( $SE$ ):
$SE = \frac{\sigma}{\sqrt{n}} = \frac{3.6}{\sqrt{35}}$
$SE \approx \frac{3.6}{5.916} \approx 0.608$

STEP 4

Calculate the z-score for the sample mean of $28\%$ :
$z = \frac{\bar{x} - \mu}{SE} = \frac{28 - 26.9}{0.608}$
$z \approx \frac{1.1}{0.608} \approx 1.809$

STEP 5

Using the z-table, find the probability that $\bar{x} > 28$ .
For $z = 1.81$ , the cumulative probability from the table is approximately $0.964$ .
Since we want $P(\bar{x} > 28)$ , we calculate:
$P(\bar{x} > 28) = 1 - P(\bar{x} \leq 28) = 1 - 0.964 = 0.036$

STEP 6

Identify the type of probability:
Since we are looking for the probability that the sample mean is greater than a certain value, this is a Right-tail probability.
The correct answers are: - Right-Tail - $P(\bar{x}>28)=0.036$

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Studdy Solution

STEP 1

1. The population mean percent allocation for bonds is μ=26.9% \mu = 26.9\% μ=26.9%.2. The population standard deviation is σ=3.6% \sigma = 3.6\% σ=3.6%.3. The sample size is n=35 n = 35 n=35.4. We are interested in the probability that the sample mean is at least 28% 28\% 28%.

STEP 2

1. Calculate the standard error of the mean.2. Calculate the z-score for the sample mean of 28% 28\% 28%.3. Determine the probability using the z-table.4. Identify the type of probability (Left-tail, Right-tail, or Interval).

STEP 3

Calculate the standard error of the mean (SE SE SE):SE=σn=3.635 SE = \frac{\sigma}{\sqrt{n}} = \frac{3.6}{\sqrt{35}} SE=n​σ​=35​3.6​SE≈3.65.916≈0.608 SE \approx \frac{3.6}{5.916} \approx 0.608 SE≈5.9163.6​≈0.608

STEP 4

Calculate the z-score for the sample mean of 28% 28\% 28%:z=xˉ−μSE=28−26.90.608 z = \frac{\bar{x} - \mu}{SE} = \frac{28 - 26.9}{0.608} z=SExˉ−μ​=0.60828−26.9​z≈1.10.608≈1.809 z \approx \frac{1.1}{0.608} \approx 1.809 z≈0.6081.1​≈1.809

STEP 5

STEP 6

Identify the type of probability:Since we are looking for the probability that the sample mean is greater than a certain value, this is a Right-tail probability. The correct answers are: - Right-Tail - P(xˉ>28)=0.036 P(\bar{x}>28)=0.036 P(xˉ>28)=0.036

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1. The population mean percent allocation for bonds is $\mu = 26.9\%$ .
2. The population standard deviation is $\sigma = 3.6\%$ .
3. The sample size is $n = 35$ .
4. We are interested in the probability that the sample mean is at least $28\%$ .

1. Calculate the standard error of the mean.
2. Calculate the z-score for the sample mean of $28\%$ .
3. Determine the probability using the z-table.
4. Identify the type of probability (Left-tail, Right-tail, or Interval).

Calculate the standard error of the mean ( $SE$ ):
$SE = \frac{\sigma}{\sqrt{n}} = \frac{3.6}{\sqrt{35}}$
$SE \approx \frac{3.6}{5.916} \approx 0.608$

Calculate the z-score for the sample mean of $28\%$ :
$z = \frac{\bar{x} - \mu}{SE} = \frac{28 - 26.9}{0.608}$
$z \approx \frac{1.1}{0.608} \approx 1.809$

Identify the type of probability:
Since we are looking for the probability that the sample mean is greater than a certain value, this is a Right-tail probability.
The correct answers are: - Right-Tail - $P(\bar{x}>28)=0.036$