Math / Data & Statistics

Questionc. Which set of data more closely resembles a normal distribution? Explain your answer.
3. In a set of data that approximate a normal distribution, the mean is 5 and the standard deviation is 2.1. (Round all answers to the nearest hundredth.) a. What percent of the values are expected to be between 2.9 and 5 ? b. What percent of the values are expected to be above 9.2 ?

Studdy Solution

STEP 1

1. We are dealing with a normal distribution, which is characterized by its mean ( $\mu$ ) and standard deviation ( $\sigma$ ).
2. The mean ( $\mu$ ) is given as $5$ and the standard deviation ( $\sigma$ ) is $2.1$ .
3. We will use the properties of the normal distribution and the standard normal distribution (z-scores) to find the required percentages.
4. We will use z-tables or the cumulative distribution function (CDF) for the standard normal distribution to find the probabilities.

STEP 2

1. Convert the given values to z-scores.
2. Use the z-scores to find the required probabilities using the z-table or CDF.
3. Interpret the results to answer the questions.

STEP 3

Convert the value $2.9$ to a z-score using the formula: $z = \frac{X - \mu}{\sigma}$ where $X = 2.9$ , $\mu = 5$ , and $\sigma = 2.1$ .
$z = \frac{2.9 - 5}{2.1} = \frac{-2.1}{2.1} = -1$

STEP 4

Convert the value $5$ to a z-score using the same formula: $z = \frac{5 - 5}{2.1} = \frac{0}{2.1} = 0$

STEP 5

Using the standard normal distribution table (z-table), find the probability corresponding to $z = -1$ .
$P(Z \leq -1) = 0.1587$

STEP 6

Using the standard normal distribution table, find the probability corresponding to $z = 0$ .
$P(Z \leq 0) = 0.5$

STEP 7

To find the probability between $z = -1$ and $z = 0$ , subtract the cumulative probability at $z = -1$ from the cumulative probability at $z = 0$ .
$P(-1 \leq Z \leq 0) = P(Z \leq 0) - P(Z \leq -1) = 0.5 - 0.1587 = 0.3413$
Therefore, the percentage of values between $2.9$ and $5$ is: $34.13\%$

STEP 8

Convert the value $9.2$ to a z-score using the formula: $z = \frac{9.2 - 5}{2.1} = \frac{4.2}{2.1} = 2$

STEP 9

Using the standard normal distribution table, find the probability corresponding to $z = 2$ .
$P(Z \leq 2) = 0.9772$

STEP 10

To find the probability of values being above $9.2$ , subtract the cumulative probability at $z = 2$ from $1$ .
$P(Z > 2) = 1 - P(Z \leq 2) = 1 - 0.9772 = 0.0228$
Therefore, the percentage of values above $9.2$ is: $2.28\%$

STEP 11

The solution to the problem is as follows: a. The percentage of values expected to be between $2.9$ and $5$ is $34.13\%$ . b. The percentage of values expected to be above $9.2$ is $2.28\%$ .

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Studdy Solution

STEP 1

STEP 2

1. Convert the given values to z-scores.
2. Use the z-scores to find the required probabilities using the z-table or CDF.
3. Interpret the results to answer the questions.

STEP 3

Convert the value $2.9$ to a z-score using the formula: $z = \frac{X - \mu}{\sigma}$ where $X = 2.9$ , $\mu = 5$ , and $\sigma = 2.1$ .
$z = \frac{2.9 - 5}{2.1} = \frac{-2.1}{2.1} = -1$

STEP 4

Convert the value $5$ to a z-score using the same formula: $z = \frac{5 - 5}{2.1} = \frac{0}{2.1} = 0$

STEP 5

Using the standard normal distribution table (z-table), find the probability corresponding to $z = -1$ .
$P(Z \leq -1) = 0.1587$

STEP 6

Using the standard normal distribution table, find the probability corresponding to $z = 0$ .
$P(Z \leq 0) = 0.5$

STEP 7

To find the probability between $z = -1$ and $z = 0$ , subtract the cumulative probability at $z = -1$ from the cumulative probability at $z = 0$ .
$P(-1 \leq Z \leq 0) = P(Z \leq 0) - P(Z \leq -1) = 0.5 - 0.1587 = 0.3413$
Therefore, the percentage of values between $2.9$ and $5$ is: $34.13\%$

STEP 8

Convert the value $9.2$ to a z-score using the formula: $z = \frac{9.2 - 5}{2.1} = \frac{4.2}{2.1} = 2$

STEP 9

Using the standard normal distribution table, find the probability corresponding to $z = 2$ .
$P(Z \leq 2) = 0.9772$

STEP 10

To find the probability of values being above $9.2$ , subtract the cumulative probability at $z = 2$ from $1$ .
$P(Z > 2) = 1 - P(Z \leq 2) = 1 - 0.9772 = 0.0228$
Therefore, the percentage of values above $9.2$ is: $2.28\%$

STEP 11

The solution to the problem is as follows: a. The percentage of values expected to be between $2.9$ and $5$ is $34.13\%$ . b. The percentage of values expected to be above $9.2$ is $2.28\%$ .

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Studdy Solution

STEP 1

STEP 2

1. Convert the given values to z-scores.2. Use the z-scores to find the required probabilities using the z-table or CDF.3. Interpret the results to answer the questions.

STEP 3

Convert the value 2.92.92.9 to a z-score using the formula: z=X−μσ z = \frac{X - \mu}{\sigma} z=σX−μ​ where X=2.9X = 2.9X=2.9, μ=5\mu = 5μ=5, and σ=2.1\sigma = 2.1σ=2.1.z=2.9−52.1=−2.12.1=−1 z = \frac{2.9 - 5}{2.1} = \frac{-2.1}{2.1} = -1 z=2.12.9−5​=2.1−2.1​=−1

STEP 4

Convert the value 555 to a z-score using the same formula: z=5−52.1=02.1=0 z = \frac{5 - 5}{2.1} = \frac{0}{2.1} = 0 z=2.15−5​=2.10​=0

STEP 5

Using the standard normal distribution table (z-table), find the probability corresponding to z=−1z = -1z=−1.P(Z≤−1)=0.1587 P(Z \leq -1) = 0.1587 P(Z≤−1)=0.1587

STEP 6

Using the standard normal distribution table, find the probability corresponding to z=0z = 0z=0.P(Z≤0)=0.5 P(Z \leq 0) = 0.5 P(Z≤0)=0.5

STEP 7

STEP 8

Convert the value 9.29.29.2 to a z-score using the formula: z=9.2−52.1=4.22.1=2 z = \frac{9.2 - 5}{2.1} = \frac{4.2}{2.1} = 2 z=2.19.2−5​=2.14.2​=2

STEP 9

Using the standard normal distribution table, find the probability corresponding to z=2z = 2z=2.P(Z≤2)=0.9772 P(Z \leq 2) = 0.9772 P(Z≤2)=0.9772

STEP 10

STEP 11

The solution to the problem is as follows: a. The percentage of values expected to be between 2.92.92.9 and 555 is 34.13%34.13\%34.13%. b. The percentage of values expected to be above 9.29.29.2 is 2.28%2.28\%2.28%.

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1. Convert the given values to z-scores.
2. Use the z-scores to find the required probabilities using the z-table or CDF.
3. Interpret the results to answer the questions.

Convert the value $2.9$ to a z-score using the formula: $z = \frac{X - \mu}{\sigma}$ where $X = 2.9$ , $\mu = 5$ , and $\sigma = 2.1$ .
$z = \frac{2.9 - 5}{2.1} = \frac{-2.1}{2.1} = -1$

Convert the value $5$ to a z-score using the same formula: $z = \frac{5 - 5}{2.1} = \frac{0}{2.1} = 0$

Using the standard normal distribution table (z-table), find the probability corresponding to $z = -1$ .
$P(Z \leq -1) = 0.1587$

Using the standard normal distribution table, find the probability corresponding to $z = 0$ .
$P(Z \leq 0) = 0.5$

Convert the value $9.2$ to a z-score using the formula: $z = \frac{9.2 - 5}{2.1} = \frac{4.2}{2.1} = 2$

Using the standard normal distribution table, find the probability corresponding to $z = 2$ .
$P(Z \leq 2) = 0.9772$

The solution to the problem is as follows: a. The percentage of values expected to be between $2.9$ and $5$ is $34.13\%$ . b. The percentage of values expected to be above $9.2$ is $2.28\%$ .