Solved on Jan 05, 2024

Find the probability that a random sample of size $n=61$ drawn from a normal distribution with $\mu=176.7$ and $\sigma=57.5$ has a mean between $166.4$ and $187.7$ .

STEP 1

Assumptions
1. The population has a normal distribution.
2. The population mean ( $\mu$ ) is 176.7.
3. The population standard deviation ( $\sigma$ ) is 57.5.
4. The sample size ( $n$ ) is 61.
5. We are looking for the probability that the sample mean ( $M$ ) is between 166.4 and 187.7.

STEP 2

Understand that the distribution of the sample mean ( $M$ ) also follows a normal distribution with mean $\mu_{M} = \mu$ and standard deviation $\sigma_{M} = \frac{\sigma}{\sqrt{n}}$ according to the Central Limit Theorem.

STEP 3

Calculate the standard deviation of the sample mean ( $\sigma_{M}$ ) using the population standard deviation ( $\sigma$ ) and the sample size ( $n$ ).
$\sigma_{M} = \frac{\sigma}{\sqrt{n}}$

STEP 4

Plug in the values for $\sigma$ and $n$ to calculate $\sigma_{M}$ .
$\sigma_{M} = \frac{57.5}{\sqrt{61}}$

STEP 5

Calculate the value of $\sigma_{M}$ .
$\sigma_{M} = \frac{57.5}{\sqrt{61}} \approx \frac{57.5}{7.81}$

STEP 6

Continue the calculation of $\sigma_{M}$ .
$\sigma_{M} \approx \frac{57.5}{7.81} \approx 7.36$

STEP 7

Now that we have $\sigma_{M}$ , we can find the z-scores for the sample mean values 166.4 and 187.7. The z-score is calculated as follows:
$z = \frac{M - \mu}{\sigma_{M}}$

STEP 8

Calculate the z-score for the lower bound of the sample mean (166.4).
$z_{lower} = \frac{166.4 - 176.7}{7.36}$

STEP 9

Calculate the value of $z_{lower}$ .
$z_{lower} = \frac{166.4 - 176.7}{7.36} \approx \frac{-10.3}{7.36} \approx -1.40$

STEP 10

Calculate the z-score for the upper bound of the sample mean (187.7).
$z_{upper} = \frac{187.7 - 176.7}{7.36}$

STEP 11

Calculate the value of $z_{upper}$ .
$z_{upper} = \frac{187.7 - 176.7}{7.36} \approx \frac{11}{7.36} \approx 1.49$

STEP 12

Use the standard normal distribution table or a calculator to find the probability corresponding to the z-scores $z_{lower}$ and $z_{upper}$ .

STEP 13

Find the probability that the z-score is less than $z_{upper}$ , which is $P(Z < 1.49)$ .

STEP 14

Find the probability that the z-score is less than $z_{lower}$ , which is $P(Z < -1.40)$ .

STEP 15

The probability that the sample mean is between 166.4 and 187.7 is the difference between the probabilities found in steps 13 and 14.
$P(166.4 < M < 187.7) = P(Z < 1.49) - P(Z < -1.40)$

STEP 16

Assuming we have the standard normal probabilities from a table or calculator, let's say $P(Z < 1.49) = 0.9319$ and $P(Z < -1.40) = 0.0808$ .

STEP 17

Subtract the probabilities to find the final probability.
$P(166.4 < M < 187.7) = 0.9319 - 0.0808$

STEP 18

Calculate the final probability.
$P(166.4 < M < 187.7) = 0.9319 - 0.0808 = 0.8511$
The probability that a sample of size $n=61$ is randomly selected with a mean between 166.4 and 187.7 is approximately 0.8511.

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Find the probability that a random sample of size $n=61$ drawn from a normal distribution with $\mu=176.7$ and $\sigma=57.5$ has a mean between $166.4$ and $187.7$ .

STEP 1

Assumptions
1. The population has a normal distribution.
2. The population mean ( $\mu$ ) is 176.7.
3. The population standard deviation ( $\sigma$ ) is 57.5.
4. The sample size ( $n$ ) is 61.
5. We are looking for the probability that the sample mean ( $M$ ) is between 166.4 and 187.7.

STEP 2

Understand that the distribution of the sample mean ( $M$ ) also follows a normal distribution with mean $\mu_{M} = \mu$ and standard deviation $\sigma_{M} = \frac{\sigma}{\sqrt{n}}$ according to the Central Limit Theorem.

STEP 3

Calculate the standard deviation of the sample mean ( $\sigma_{M}$ ) using the population standard deviation ( $\sigma$ ) and the sample size ( $n$ ).
$\sigma_{M} = \frac{\sigma}{\sqrt{n}}$

STEP 4

Plug in the values for $\sigma$ and $n$ to calculate $\sigma_{M}$ .
$\sigma_{M} = \frac{57.5}{\sqrt{61}}$

STEP 5

Calculate the value of $\sigma_{M}$ .
$\sigma_{M} = \frac{57.5}{\sqrt{61}} \approx \frac{57.5}{7.81}$

STEP 6

Continue the calculation of $\sigma_{M}$ .
$\sigma_{M} \approx \frac{57.5}{7.81} \approx 7.36$

STEP 7

Now that we have $\sigma_{M}$ , we can find the z-scores for the sample mean values 166.4 and 187.7. The z-score is calculated as follows:
$z = \frac{M - \mu}{\sigma_{M}}$

STEP 8

Calculate the z-score for the lower bound of the sample mean (166.4).
$z_{lower} = \frac{166.4 - 176.7}{7.36}$

STEP 9

Calculate the value of $z_{lower}$ .
$z_{lower} = \frac{166.4 - 176.7}{7.36} \approx \frac{-10.3}{7.36} \approx -1.40$

STEP 10

Calculate the z-score for the upper bound of the sample mean (187.7).
$z_{upper} = \frac{187.7 - 176.7}{7.36}$

STEP 11

Calculate the value of $z_{upper}$ .
$z_{upper} = \frac{187.7 - 176.7}{7.36} \approx \frac{11}{7.36} \approx 1.49$

STEP 12

Use the standard normal distribution table or a calculator to find the probability corresponding to the z-scores $z_{lower}$ and $z_{upper}$ .

STEP 13

Find the probability that the z-score is less than $z_{upper}$ , which is $P(Z < 1.49)$ .

STEP 14

Find the probability that the z-score is less than $z_{lower}$ , which is $P(Z < -1.40)$ .

STEP 15

The probability that the sample mean is between 166.4 and 187.7 is the difference between the probabilities found in steps 13 and 14.
$P(166.4 < M < 187.7) = P(Z < 1.49) - P(Z < -1.40)$

STEP 16

Assuming we have the standard normal probabilities from a table or calculator, let's say $P(Z < 1.49) = 0.9319$ and $P(Z < -1.40) = 0.0808$ .

STEP 17

Subtract the probabilities to find the final probability.
$P(166.4 < M < 187.7) = 0.9319 - 0.0808$

STEP 18

Calculate the final probability.
$P(166.4 < M < 187.7) = 0.9319 - 0.0808 = 0.8511$
The probability that a sample of size $n=61$ is randomly selected with a mean between 166.4 and 187.7 is approximately 0.8511.

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Find the probability that a random sample of size n=61n=61n=61 drawn from a normal distribution with μ=176.7\mu=176.7μ=176.7 and σ=57.5\sigma=57.5σ=57.5 has a mean between 166.4166.4166.4 and 187.7187.7187.7.

STEP 1

Assumptions1. The population has a normal distribution.2. The population mean (μ\muμ) is 176.7.3. The population standard deviation (σ\sigmaσ) is 57.5.4. The sample size (nnn) is 61.5. We are looking for the probability that the sample mean (MMM) is between 166.4 and 187.7.

STEP 2

Understand that the distribution of the sample mean (MMM) also follows a normal distribution with mean μM=μ\mu_{M} = \muμM​=μ and standard deviation σM=σn\sigma_{M} = \frac{\sigma}{\sqrt{n}}σM​=n​σ​ according to the Central Limit Theorem.

STEP 3

Calculate the standard deviation of the sample mean (σM\sigma_{M}σM​) using the population standard deviation (σ\sigmaσ) and the sample size (nnn).σM=σn\sigma_{M} = \frac{\sigma}{\sqrt{n}}σM​=n​σ​

STEP 4

Plug in the values for σ\sigmaσ and nnn to calculate σM\sigma_{M}σM​.σM=57.561\sigma_{M} = \frac{57.5}{\sqrt{61}}σM​=61​57.5​

STEP 5

Calculate the value of σM\sigma_{M}σM​.σM=57.561≈57.57.81\sigma_{M} = \frac{57.5}{\sqrt{61}} \approx \frac{57.5}{7.81}σM​=61​57.5​≈7.8157.5​

STEP 6

Continue the calculation of σM\sigma_{M}σM​.σM≈57.57.81≈7.36\sigma_{M} \approx \frac{57.5}{7.81} \approx 7.36σM​≈7.8157.5​≈7.36

STEP 7

Now that we have σM\sigma_{M}σM​, we can find the z-scores for the sample mean values 166.4 and 187.7. The z-score is calculated as follows:z=M−μσMz = \frac{M - \mu}{\sigma_{M}}z=σM​M−μ​

STEP 8

Calculate the z-score for the lower bound of the sample mean (166.4).zlower=166.4−176.77.36z_{lower} = \frac{166.4 - 176.7}{7.36}zlower​=7.36166.4−176.7​

STEP 9

Calculate the value of zlowerz_{lower}zlower​.zlower=166.4−176.77.36≈−10.37.36≈−1.40z_{lower} = \frac{166.4 - 176.7}{7.36} \approx \frac{-10.3}{7.36} \approx -1.40zlower​=7.36166.4−176.7​≈7.36−10.3​≈−1.40

STEP 10

Calculate the z-score for the upper bound of the sample mean (187.7).zupper=187.7−176.77.36z_{upper} = \frac{187.7 - 176.7}{7.36}zupper​=7.36187.7−176.7​

STEP 11

Calculate the value of zupperz_{upper}zupper​.zupper=187.7−176.77.36≈117.36≈1.49z_{upper} = \frac{187.7 - 176.7}{7.36} \approx \frac{11}{7.36} \approx 1.49zupper​=7.36187.7−176.7​≈7.3611​≈1.49

STEP 12

Use the standard normal distribution table or a calculator to find the probability corresponding to the z-scores zlowerz_{lower}zlower​ and zupperz_{upper}zupper​.

STEP 13

Find the probability that the z-score is less than zupperz_{upper}zupper​, which is P(Z<1.49)P(Z < 1.49)P(Z<1.49).

STEP 14

Find the probability that the z-score is less than zlowerz_{lower}zlower​, which is P(Z<−1.40)P(Z < -1.40)P(Z<−1.40).

STEP 15

The probability that the sample mean is between 166.4 and 187.7 is the difference between the probabilities found in steps 13 and 14.P(166.4<M<187.7)=P(Z<1.49)−P(Z<−1.40)P(166.4 < M < 187.7) = P(Z < 1.49) - P(Z < -1.40)P(166.4<M<187.7)=P(Z<1.49)−P(Z<−1.40)

STEP 16

Assuming we have the standard normal probabilities from a table or calculator, let's say P(Z<1.49)=0.9319P(Z < 1.49) = 0.9319P(Z<1.49)=0.9319 and P(Z<−1.40)=0.0808P(Z < -1.40) = 0.0808P(Z<−1.40)=0.0808.

STEP 17

Subtract the probabilities to find the final probability.P(166.4<M<187.7)=0.9319−0.0808P(166.4 < M < 187.7) = 0.9319 - 0.0808P(166.4<M<187.7)=0.9319−0.0808

STEP 18

Calculate the final probability.P(166.4<M<187.7)=0.9319−0.0808=0.8511P(166.4 < M < 187.7) = 0.9319 - 0.0808 = 0.8511P(166.4<M<187.7)=0.9319−0.0808=0.8511The probability that a sample of size n=61n=61n=61 is randomly selected with a mean between 166.4 and 187.7 is approximately 0.8511.

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Find the probability that a random sample of size $n=61$ drawn from a normal distribution with $\mu=176.7$ and $\sigma=57.5$ has a mean between $166.4$ and $187.7$ .

Assumptions
1. The population has a normal distribution.
2. The population mean ( $\mu$ ) is 176.7.
3. The population standard deviation ( $\sigma$ ) is 57.5.
4. The sample size ( $n$ ) is 61.
5. We are looking for the probability that the sample mean ( $M$ ) is between 166.4 and 187.7.

Understand that the distribution of the sample mean ( $M$ ) also follows a normal distribution with mean $\mu_{M} = \mu$ and standard deviation $\sigma_{M} = \frac{\sigma}{\sqrt{n}}$ according to the Central Limit Theorem.

Calculate the standard deviation of the sample mean ( $\sigma_{M}$ ) using the population standard deviation ( $\sigma$ ) and the sample size ( $n$ ).
$\sigma_{M} = \frac{\sigma}{\sqrt{n}}$

Plug in the values for $\sigma$ and $n$ to calculate $\sigma_{M}$ .
$\sigma_{M} = \frac{57.5}{\sqrt{61}}$

Calculate the value of $\sigma_{M}$ .
$\sigma_{M} = \frac{57.5}{\sqrt{61}} \approx \frac{57.5}{7.81}$

Continue the calculation of $\sigma_{M}$ .
$\sigma_{M} \approx \frac{57.5}{7.81} \approx 7.36$

Now that we have $\sigma_{M}$ , we can find the z-scores for the sample mean values 166.4 and 187.7. The z-score is calculated as follows:
$z = \frac{M - \mu}{\sigma_{M}}$

Calculate the z-score for the lower bound of the sample mean (166.4).
$z_{lower} = \frac{166.4 - 176.7}{7.36}$

Calculate the value of $z_{lower}$ .
$z_{lower} = \frac{166.4 - 176.7}{7.36} \approx \frac{-10.3}{7.36} \approx -1.40$

Calculate the z-score for the upper bound of the sample mean (187.7).
$z_{upper} = \frac{187.7 - 176.7}{7.36}$

Calculate the value of $z_{upper}$ .
$z_{upper} = \frac{187.7 - 176.7}{7.36} \approx \frac{11}{7.36} \approx 1.49$

Use the standard normal distribution table or a calculator to find the probability corresponding to the z-scores $z_{lower}$ and $z_{upper}$ .

Find the probability that the z-score is less than $z_{upper}$ , which is $P(Z < 1.49)$ .

Find the probability that the z-score is less than $z_{lower}$ , which is $P(Z < -1.40)$ .

The probability that the sample mean is between 166.4 and 187.7 is the difference between the probabilities found in steps 13 and 14.
$P(166.4 < M < 187.7) = P(Z < 1.49) - P(Z < -1.40)$

Assuming we have the standard normal probabilities from a table or calculator, let's say $P(Z < 1.49) = 0.9319$ and $P(Z < -1.40) = 0.0808$ .

Subtract the probabilities to find the final probability.
$P(166.4 < M < 187.7) = 0.9319 - 0.0808$

Calculate the final probability.
$P(166.4 < M < 187.7) = 0.9319 - 0.0808 = 0.8511$
The probability that a sample of size $n=61$ is randomly selected with a mean between 166.4 and 187.7 is approximately 0.8511.