Math / Data & Statistics

QuestionA sample of size 23 will be drawn from a population with mean 6 and standard deviation 5 . (a) Is it appropriate to use the normal distribution to find probabilities for $\bar{x}$ ? (b) If appropriate find the probability that $\bar{x}$ will be greater than 4. (c) If appropriate find the $20^{\text {th }}$ percentile of $\bar{x}$ . It is appropriate to use the normal distribution to find probabilities for $\bar{x}$ . The probability that $\bar{x}$ will be greater than 4 is $\square$ The $20^{\text {th }}$ percentile of $\bar{x}$ is $\square$ . It is not appropriate to use the normal distribution to find probabilities for $\bar{x}$ .

Studdy Solution

STEP 1

What is this asking? We're checking if it's okay to use the normal distribution for sample means, and if so, we'll find the probability that a sample mean is greater than 4, and the 20th percentile of the sample means. Watch out! Remember, we need to use the standard deviation of the sample mean, not the population standard deviation, when working with sample means!

STEP 2

1. Check Normality
2. Calculate Probability
3. Find Percentile

STEP 3

The problem states the sample size is $\text{n} = 23$ .
Since the sample size is greater than 30, we can assume the sampling distribution of the sample mean is approximately normal due to the Central Limit Theorem.

STEP 4

The standard deviation of the sample mean, denoted by $\sigma_{\bar{x}}$ , is calculated by dividing the population standard deviation, $\sigma$ , by the square root of the sample size, $n$ .
So, $\sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}}$ .

STEP 5

We know that $\sigma = 5$ and $n = 23$ , so $\sigma_{\bar{x}} = \frac{5}{\sqrt{23}} \approx 1.04$ .

STEP 6

The z-score is calculated as $z = \frac{\bar{x} - \mu}{\sigma_{\bar{x}}}$ , where $\bar{x}$ is the sample mean we're interested in, $\mu$ is the population mean, and $\sigma_{\bar{x}}$ is the standard deviation of the sample mean.
We want to find the probability that $\bar{x}$ is greater than 4, so our $\bar{x}$ is 4.
The population mean $\mu$ is 6.
Plugging in the values, we get $z = \frac{4 - 6}{1.04} \approx -1.92$ .

STEP 7

Using a z-table or calculator, we find that the probability of a z-score being greater than -1.92 is approximately 0.973.

STEP 8

The 20th percentile corresponds to a cumulative probability of 0.20.
Using a z-table or calculator, we find the z-score corresponding to a cumulative probability of 0.20 is approximately -0.84.

STEP 9

We can use the formula $\bar{x} = \mu + z \cdot \sigma_{\bar{x}}$ to find the 20th percentile of $\bar{x}$ .
Plugging in the values, we get $\bar{x} = 6 + (-0.84) \cdot 1.04 \approx 5.13$ .

STEP 10

(a) Yes, it is appropriate to use the normal distribution. (b) The probability that $\bar{x}$ will be greater than 4 is approximately 0.973. (c) The 20th percentile of $\bar{x}$ is approximately 5.13.

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Studdy Solution

STEP 1

STEP 2

1. Check Normality
2. Calculate Probability
3. Find Percentile

STEP 3

The problem states the sample size is $\text{n} = 23$ .
Since the sample size is greater than 30, we can assume the sampling distribution of the sample mean is approximately normal due to the Central Limit Theorem.

STEP 4

The standard deviation of the sample mean, denoted by $\sigma_{\bar{x}}$ , is calculated by dividing the population standard deviation, $\sigma$ , by the square root of the sample size, $n$ .
So, $\sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}}$ .

STEP 5

We know that $\sigma = 5$ and $n = 23$ , so $\sigma_{\bar{x}} = \frac{5}{\sqrt{23}} \approx 1.04$ .

STEP 6

STEP 7

Using a z-table or calculator, we find that the probability of a z-score being greater than -1.92 is approximately 0.973.

STEP 8

The 20th percentile corresponds to a cumulative probability of 0.20.
Using a z-table or calculator, we find the z-score corresponding to a cumulative probability of 0.20 is approximately -0.84.

STEP 9

We can use the formula $\bar{x} = \mu + z \cdot \sigma_{\bar{x}}$ to find the 20th percentile of $\bar{x}$ .
Plugging in the values, we get $\bar{x} = 6 + (-0.84) \cdot 1.04 \approx 5.13$ .

STEP 10

(a) Yes, it is appropriate to use the normal distribution. (b) The probability that $\bar{x}$ will be greater than 4 is approximately 0.973. (c) The 20th percentile of $\bar{x}$ is approximately 5.13.

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Studdy Solution

STEP 1

STEP 2

1. Check Normality2. Calculate Probability3. Find Percentile

STEP 3

The problem states the sample size is n=23\text{n} = 23n=23.Since the sample size is greater than 30, we can assume the sampling distribution of the sample mean is approximately normal due to the Central Limit Theorem.

STEP 4

The standard deviation of the sample mean, denoted by σxˉ\sigma_{\bar{x}}σxˉ​, is calculated by dividing the population standard deviation, σ\sigmaσ, by the square root of the sample size, nnn.So, σxˉ=σn\sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}}σxˉ​=n​σ​.

STEP 5

We know that σ=5\sigma = 5σ=5 and n=23n = 23n=23, so σxˉ=523≈1.04\sigma_{\bar{x}} = \frac{5}{\sqrt{23}} \approx 1.04σxˉ​=23​5​≈1.04.

STEP 6

STEP 7

Using a z-table or calculator, we find that the probability of a z-score being greater than **-1.92** is approximately **0.973**.

STEP 8

The 20th percentile corresponds to a cumulative probability of **0.20**.Using a z-table or calculator, we find the z-score corresponding to a cumulative probability of **0.20** is approximately **-0.84**.

STEP 9

We can use the formula xˉ=μ+z⋅σxˉ\bar{x} = \mu + z \cdot \sigma_{\bar{x}}xˉ=μ+z⋅σxˉ​ to find the 20th percentile of xˉ\bar{x}xˉ.Plugging in the values, we get xˉ=6+(−0.84)⋅1.04≈5.13\bar{x} = 6 + (-0.84) \cdot 1.04 \approx 5.13xˉ=6+(−0.84)⋅1.04≈5.13.

STEP 10

(a) Yes, it is appropriate to use the normal distribution. (b) The probability that xˉ\bar{x}xˉ will be greater than 4 is approximately **0.973**. (c) The 20th percentile of xˉ\bar{x}xˉ is approximately **5.13**.

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1. Check Normality
2. Calculate Probability
3. Find Percentile

The problem states the sample size is $\text{n} = 23$ .
Since the sample size is greater than 30, we can assume the sampling distribution of the sample mean is approximately normal due to the Central Limit Theorem.

The standard deviation of the sample mean, denoted by $\sigma_{\bar{x}}$ , is calculated by dividing the population standard deviation, $\sigma$ , by the square root of the sample size, $n$ .
So, $\sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}}$ .

We know that $\sigma = 5$ and $n = 23$ , so $\sigma_{\bar{x}} = \frac{5}{\sqrt{23}} \approx 1.04$ .

Using a z-table or calculator, we find that the probability of a z-score being greater than -1.92 is approximately 0.973.

The 20th percentile corresponds to a cumulative probability of 0.20.
Using a z-table or calculator, we find the z-score corresponding to a cumulative probability of 0.20 is approximately -0.84.

We can use the formula $\bar{x} = \mu + z \cdot \sigma_{\bar{x}}$ to find the 20th percentile of $\bar{x}$ .
Plugging in the values, we get $\bar{x} = 6 + (-0.84) \cdot 1.04 \approx 5.13$ .

(a) Yes, it is appropriate to use the normal distribution. (b) The probability that $\bar{x}$ will be greater than 4 is approximately 0.973. (c) The 20th percentile of $\bar{x}$ is approximately 5.13.