Math / Data & Statistics

Question $\checkmark 1$ 2 $\checkmark 3$ 6 7 8 10 $\checkmark 11$ 12 Español
TV sets: According to the Nielsen Company, the mean number of TV sets in a U.S. household was 2.24. Assume the standard deviation is 1.2 . A sample of 95 households is drawn.
Part 1 of 5 (a) What is the probability that the sample mean number of TV sets is greater than 2? Round your answer to at least four decimal places.
The probability that the sample mean number of TV sets is greater than 2 is 0.9744 .
Part: $1 / 5$
Part 2 of 5 (b) What is the probability that the sample mean number of TV sets is between 2.5 and 3? Round your answer to at least four decimal places.
The probability that the sample mean number of TV sets is between 2.5 and 3 is $\square$ . Skip Part Check Save For Later Submit Assignment

Studdy Solution

STEP 1

What is this asking? What's the chance that a group of 95 households has, on average, between 2.5 and 3 TVs? Watch out! Don't mix up the number of households with the number of TVs!
Also, remember the difference between population parameters and sample statistics.

STEP 2

1. Calculate the standard error.
2. Calculate the z-scores.
3. Find the probability between the z-scores.

STEP 3

We're given that the population standard deviation $\sigma$ is $1.2$ and the sample size $n$ is 95.
The standard error $\sigma_{\bar{x}}$ measures the variability of the sample means.

STEP 4

To calculate the standard error, we divide the population standard deviation by the square root of the sample size: $\sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}}$

STEP 5

Let's plug in our values: $\sigma_{\bar{x}} = \frac{1.2}{\sqrt{95}} \approx 0.123$ So, our standard error is approximately 0.123.

STEP 6

We want to find the probability that the sample mean $\bar{x}$ is between 2.5 and 3.
To do this, we'll convert these sample mean values to z-scores.
A z-score tells us how many standard errors a value is away from the population mean.

STEP 7

The formula for the z-score is: $z = \frac{\bar{x} - \mu}{\sigma_{\bar{x}}}$ where $\mu$ is the population mean, which is given as 2.24.

STEP 8

For $\bar{x} = 2.5$ : $z_1 = \frac{2.5 - 2.24}{0.123} \approx 2.11$

STEP 9

For $\bar{x} = 3$ : $z_2 = \frac{3 - 2.24}{0.123} \approx 6.18$

STEP 10

Now, we want to find the probability that the z-score is between $z_1 \approx 2.11$ and $z_2 \approx 6.18$ .
We can look this up in a z-table or use a calculator.

STEP 11

The probability corresponding to $z_1 \approx 2.11$ is approximately $0.9826$ .
Since $z_2 \approx 6.18$ is so large, the probability to the left of it is essentially 1.

STEP 12

To find the probability between the two z-scores, we subtract the smaller probability from the larger probability: $P(2.11 < z < 6.18) \approx 1 - 0.9826 = 0.0174$

STEP 13

The probability that the sample mean number of TV sets is between 2.5 and 3 is approximately 0.0174.

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Studdy Solution

STEP 1

What is this asking? What's the chance that a group of 95 households has, on average, between 2.5 and 3 TVs? Watch out! Don't mix up the number of households with the number of TVs!
Also, remember the difference between population parameters and sample statistics.

STEP 2

1. Calculate the standard error.
2. Calculate the z-scores.
3. Find the probability between the z-scores.

STEP 3

We're given that the population standard deviation $\sigma$ is $1.2$ and the sample size $n$ is 95.
The standard error $\sigma_{\bar{x}}$ measures the variability of the sample means.

STEP 4

To calculate the standard error, we divide the population standard deviation by the square root of the sample size: $\sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}}$

STEP 5

Let's plug in our values: $\sigma_{\bar{x}} = \frac{1.2}{\sqrt{95}} \approx 0.123$ So, our standard error is approximately 0.123.

STEP 6

We want to find the probability that the sample mean $\bar{x}$ is between 2.5 and 3.
To do this, we'll convert these sample mean values to z-scores.
A z-score tells us how many standard errors a value is away from the population mean.

STEP 7

The formula for the z-score is: $z = \frac{\bar{x} - \mu}{\sigma_{\bar{x}}}$ where $\mu$ is the population mean, which is given as 2.24.

STEP 8

For $\bar{x} = 2.5$ : $z_1 = \frac{2.5 - 2.24}{0.123} \approx 2.11$

STEP 9

For $\bar{x} = 3$ : $z_2 = \frac{3 - 2.24}{0.123} \approx 6.18$

STEP 10

Now, we want to find the probability that the z-score is between $z_1 \approx 2.11$ and $z_2 \approx 6.18$ .
We can look this up in a z-table or use a calculator.

STEP 11

The probability corresponding to $z_1 \approx 2.11$ is approximately $0.9826$ .
Since $z_2 \approx 6.18$ is so large, the probability to the left of it is essentially 1.

STEP 12

To find the probability between the two z-scores, we subtract the smaller probability from the larger probability: $P(2.11 < z < 6.18) \approx 1 - 0.9826 = 0.0174$

STEP 13

The probability that the sample mean number of TV sets is between 2.5 and 3 is approximately 0.0174.

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Studdy Solution

STEP 1

What is this asking? What's the chance that a group of 95 households has, on average, between 2.5 and 3 TVs? Watch out! Don't mix up the number of households with the number of TVs!Also, remember the difference between population parameters and sample statistics.

STEP 2

1. Calculate the standard error.2. Calculate the z-scores.3. Find the probability between the z-scores.

STEP 3

We're given that the **population standard deviation** σ\sigmaσ is 1.21.21.2 and the **sample size** nnn is **95**.The **standard error** σxˉ\sigma_{\bar{x}}σxˉ​ measures the variability of the sample means.

STEP 4

To **calculate the standard error**, we divide the population standard deviation by the square root of the sample size: σxˉ=σn \sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}} σxˉ​=n​σ​

STEP 5

Let's plug in our values: σxˉ=1.295≈0.123 \sigma_{\bar{x}} = \frac{1.2}{\sqrt{95}} \approx 0.123 σxˉ​=95​1.2​≈0.123 So, our **standard error** is approximately **0.123**.

STEP 6

We want to find the probability that the sample mean xˉ\bar{x}xˉ is between 2.5 and 3.To do this, we'll convert these sample mean values to **z-scores**.A z-score tells us how many standard errors a value is away from the population mean.

STEP 7

The formula for the **z-score** is: z=xˉ−μσxˉ z = \frac{\bar{x} - \mu}{\sigma_{\bar{x}}} z=σxˉ​xˉ−μ​ where μ\muμ is the **population mean**, which is given as **2.24**.

STEP 8

For xˉ=2.5\bar{x} = 2.5xˉ=2.5: z1=2.5−2.240.123≈2.11 z_1 = \frac{2.5 - 2.24}{0.123} \approx 2.11 z1​=0.1232.5−2.24​≈2.11

STEP 9

For xˉ=3\bar{x} = 3xˉ=3: z2=3−2.240.123≈6.18 z_2 = \frac{3 - 2.24}{0.123} \approx 6.18 z2​=0.1233−2.24​≈6.18

STEP 10

Now, we want to find the probability that the z-score is between z1≈2.11z_1 \approx 2.11z1​≈2.11 and z2≈6.18z_2 \approx 6.18z2​≈6.18.We can look this up in a **z-table** or use a calculator.

STEP 11

The probability corresponding to z1≈2.11z_1 \approx 2.11z1​≈2.11 is approximately 0.98260.98260.9826.Since z2≈6.18z_2 \approx 6.18z2​≈6.18 is so large, the probability to the left of it is essentially 1.

STEP 12

To find the probability between the two z-scores, we subtract the smaller probability from the larger probability: P(2.11<z<6.18)≈1−0.9826=0.0174 P(2.11 < z < 6.18) \approx 1 - 0.9826 = 0.0174 P(2.11<z<6.18)≈1−0.9826=0.0174

STEP 13

The probability that the sample mean number of TV sets is between 2.5 and 3 is approximately **0.0174**.

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Start learning now

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What is this asking? What's the chance that a group of 95 households has, on average, between 2.5 and 3 TVs? Watch out! Don't mix up the number of households with the number of TVs!
Also, remember the difference between population parameters and sample statistics.

1. Calculate the standard error.
2. Calculate the z-scores.
3. Find the probability between the z-scores.

We're given that the population standard deviation $\sigma$ is $1.2$ and the sample size $n$ is 95.
The standard error $\sigma_{\bar{x}}$ measures the variability of the sample means.

To calculate the standard error, we divide the population standard deviation by the square root of the sample size: $\sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}}$

Let's plug in our values: $\sigma_{\bar{x}} = \frac{1.2}{\sqrt{95}} \approx 0.123$ So, our standard error is approximately 0.123.

We want to find the probability that the sample mean $\bar{x}$ is between 2.5 and 3.
To do this, we'll convert these sample mean values to z-scores.
A z-score tells us how many standard errors a value is away from the population mean.

The formula for the z-score is: $z = \frac{\bar{x} - \mu}{\sigma_{\bar{x}}}$ where $\mu$ is the population mean, which is given as 2.24.

For $\bar{x} = 2.5$ : $z_1 = \frac{2.5 - 2.24}{0.123} \approx 2.11$

For $\bar{x} = 3$ : $z_2 = \frac{3 - 2.24}{0.123} \approx 6.18$

Now, we want to find the probability that the z-score is between $z_1 \approx 2.11$ and $z_2 \approx 6.18$ .
We can look this up in a z-table or use a calculator.

The probability corresponding to $z_1 \approx 2.11$ is approximately $0.9826$ .
Since $z_2 \approx 6.18$ is so large, the probability to the left of it is essentially 1.

To find the probability between the two z-scores, we subtract the smaller probability from the larger probability: $P(2.11 < z < 6.18) \approx 1 - 0.9826 = 0.0174$

The probability that the sample mean number of TV sets is between 2.5 and 3 is approximately 0.0174.