Solved on Dec 05, 2023

An elevator has a capacity of 2385 lb for 15 passengers. Find the probability that 15 adult males with $\mu = 165$ lb, $\sigma = 32$ lb have a mean weight > 159 lb, indicating an overload. Does the elevator appear safe?

STEP 1

Assumptions
1. The maximum capacity of the elevator is 2385 lb for 15 passengers.
2. The mean weight of adult male passengers is 165 lb.
3. The standard deviation of the weight of adult male passengers is 32 lb.
4. The weights of males are normally distributed.
5. We are looking for the probability that the mean weight of 15 passengers exceeds 159 lb.

STEP 2

First, we need to find the mean weight per passenger that the elevator can carry without being overloaded.
$\text{Maximum mean weight per passenger} = \frac{\text{Maximum capacity}}{\text{Number of passengers}}$

STEP 3

Now, plug in the given values for the maximum capacity and the number of passengers to calculate the maximum mean weight per passenger.
$\text{Maximum mean weight per passenger} = \frac{2385}{15}$

STEP 4

Calculate the maximum mean weight per passenger.
$\text{Maximum mean weight per passenger} = \frac{2385}{15} = 159 \text{ lb}$

STEP 5

We need to find the probability that the mean weight of 15 passengers is greater than 159 lb. To do this, we use the standard normal distribution and the concept of the sampling distribution of the sample mean.

STEP 6

First, we calculate the standard error of the mean (SEM), which is the standard deviation of the sampling distribution of the sample mean.
$SEM = \frac{\sigma}{\sqrt{n}}$
where $\sigma$ is the population standard deviation and $n$ is the sample size.

STEP 7

Now, plug in the given values for the population standard deviation and the sample size to calculate the SEM.
$SEM = \frac{32}{\sqrt{15}}$

STEP 8

Calculate the SEM.
$SEM = \frac{32}{\sqrt{15}} \approx 8.26 \text{ lb}$

STEP 9

Next, we calculate the z-score, which is the number of standard errors the sample mean is away from the population mean.
$z = \frac{\bar{x} - \mu}{SEM}$
where $\bar{x}$ is the sample mean, $\mu$ is the population mean, and SEM is the standard error of the mean.

STEP 10

Here, we are interested in the probability that the sample mean is greater than 159 lb. So, $\bar{x} = 159$ lb and $\mu = 165$ lb.
$z = \frac{159 - 165}{8.26}$

STEP 11

Calculate the z-score.
$z = \frac{159 - 165}{8.26} \approx -0.73$

STEP 12

To find the probability that the sample mean is greater than 159 lb, we look up the z-score in the standard normal distribution table, or we can use a calculator or software that provides the cumulative distribution function (CDF) for the standard normal distribution.

STEP 13

The probability we want is the area to the right of the z-score in the standard normal distribution. This is given by $1 - \text{CDF}(z)$ .
$P(\bar{x} > 159) = 1 - \text{CDF}(-0.73)$

STEP 14

Using standard normal distribution tables or software, we find the CDF for $z = -0.73$ .

STEP 15

The CDF for $z = -0.73$ is approximately 0.2327 (this value may vary slightly depending on the table or software used).

STEP 16

Now, calculate the probability that the sample mean is greater than 159 lb.
$P(\bar{x} > 159) = 1 - 0.2327$

STEP 17

Calculate the final probability.
$P(\bar{x} > 159) = 1 - 0.2327 \approx 0.7673$

STEP 18

The probability that the elevator is overloaded with 15 adult male passengers having a mean weight greater than 159 lb is approximately 0.7673, or 76.73%.

STEP 19

To determine if the elevator appears to be safe, we consider the calculated probability. A probability of 76.73% indicates that there is a high likelihood that the elevator will be overloaded if it is filled with 15 adult male passengers. This suggests that the elevator may not be safe under these conditions.
The probability that the elevator is overloaded is approximately 76.73%, which indicates that the elevator may not be safe for 15 adult male passengers with the given weight distribution.

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An elevator has a capacity of 2385 lb for 15 passengers. Find the probability that 15 adult males with $\mu = 165$ lb, $\sigma = 32$ lb have a mean weight > 159 lb, indicating an overload. Does the elevator appear safe?

STEP 1

STEP 2

First, we need to find the mean weight per passenger that the elevator can carry without being overloaded.
$\text{Maximum mean weight per passenger} = \frac{\text{Maximum capacity}}{\text{Number of passengers}}$

STEP 3

Now, plug in the given values for the maximum capacity and the number of passengers to calculate the maximum mean weight per passenger.
$\text{Maximum mean weight per passenger} = \frac{2385}{15}$

STEP 4

Calculate the maximum mean weight per passenger.
$\text{Maximum mean weight per passenger} = \frac{2385}{15} = 159 \text{ lb}$

STEP 5

We need to find the probability that the mean weight of 15 passengers is greater than 159 lb. To do this, we use the standard normal distribution and the concept of the sampling distribution of the sample mean.

STEP 6

First, we calculate the standard error of the mean (SEM), which is the standard deviation of the sampling distribution of the sample mean.
$SEM = \frac{\sigma}{\sqrt{n}}$
where $\sigma$ is the population standard deviation and $n$ is the sample size.

STEP 7

Now, plug in the given values for the population standard deviation and the sample size to calculate the SEM.
$SEM = \frac{32}{\sqrt{15}}$

STEP 8

Calculate the SEM.
$SEM = \frac{32}{\sqrt{15}} \approx 8.26 \text{ lb}$

STEP 9

Next, we calculate the z-score, which is the number of standard errors the sample mean is away from the population mean.
$z = \frac{\bar{x} - \mu}{SEM}$
where $\bar{x}$ is the sample mean, $\mu$ is the population mean, and SEM is the standard error of the mean.

STEP 10

Here, we are interested in the probability that the sample mean is greater than 159 lb. So, $\bar{x} = 159$ lb and $\mu = 165$ lb.
$z = \frac{159 - 165}{8.26}$

STEP 11

Calculate the z-score.
$z = \frac{159 - 165}{8.26} \approx -0.73$

STEP 12

To find the probability that the sample mean is greater than 159 lb, we look up the z-score in the standard normal distribution table, or we can use a calculator or software that provides the cumulative distribution function (CDF) for the standard normal distribution.

STEP 13

The probability we want is the area to the right of the z-score in the standard normal distribution. This is given by $1 - \text{CDF}(z)$ .
$P(\bar{x} > 159) = 1 - \text{CDF}(-0.73)$

STEP 14

Using standard normal distribution tables or software, we find the CDF for $z = -0.73$ .

STEP 15

The CDF for $z = -0.73$ is approximately 0.2327 (this value may vary slightly depending on the table or software used).

STEP 16

Now, calculate the probability that the sample mean is greater than 159 lb.
$P(\bar{x} > 159) = 1 - 0.2327$

STEP 17

Calculate the final probability.
$P(\bar{x} > 159) = 1 - 0.2327 \approx 0.7673$

STEP 18

The probability that the elevator is overloaded with 15 adult male passengers having a mean weight greater than 159 lb is approximately 0.7673, or 76.73%.

STEP 19

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An elevator has a capacity of 2385 lb for 15 passengers. Find the probability that 15 adult males with μ=165\mu = 165μ=165 lb, σ=32\sigma = 32σ=32 lb have a mean weight > 159 lb, indicating an overload. Does the elevator appear safe?

STEP 1

STEP 2

STEP 3

Now, plug in the given values for the maximum capacity and the number of passengers to calculate the maximum mean weight per passenger.Maximum mean weight per passenger=238515\text{Maximum mean weight per passenger} = \frac{2385}{15}Maximum mean weight per passenger=152385​

STEP 4

Calculate the maximum mean weight per passenger.Maximum mean weight per passenger=238515=159 lb\text{Maximum mean weight per passenger} = \frac{2385}{15} = 159 \text{ lb}Maximum mean weight per passenger=152385​=159 lb

STEP 5

We need to find the probability that the mean weight of 15 passengers is greater than 159 lb. To do this, we use the standard normal distribution and the concept of the sampling distribution of the sample mean.

STEP 6

First, we calculate the standard error of the mean (SEM), which is the standard deviation of the sampling distribution of the sample mean.SEM=σnSEM = \frac{\sigma}{\sqrt{n}}SEM=n​σ​where σ\sigmaσ is the population standard deviation and nnn is the sample size.

STEP 7

Now, plug in the given values for the population standard deviation and the sample size to calculate the SEM.SEM=3215SEM = \frac{32}{\sqrt{15}}SEM=15​32​

STEP 8

Calculate the SEM.SEM=3215≈8.26 lbSEM = \frac{32}{\sqrt{15}} \approx 8.26 \text{ lb}SEM=15​32​≈8.26 lb

STEP 9

Next, we calculate the z-score, which is the number of standard errors the sample mean is away from the population mean.z=xˉ−μSEMz = \frac{\bar{x} - \mu}{SEM}z=SEMxˉ−μ​where xˉ\bar{x}xˉ is the sample mean, μ\muμ is the population mean, and SEM is the standard error of the mean.

STEP 10

Here, we are interested in the probability that the sample mean is greater than 159 lb. So, xˉ=159\bar{x} = 159xˉ=159 lb and μ=165\mu = 165μ=165 lb.z=159−1658.26z = \frac{159 - 165}{8.26}z=8.26159−165​

STEP 11

Calculate the z-score.z=159−1658.26≈−0.73z = \frac{159 - 165}{8.26} \approx -0.73z=8.26159−165​≈−0.73

STEP 12

To find the probability that the sample mean is greater than 159 lb, we look up the z-score in the standard normal distribution table, or we can use a calculator or software that provides the cumulative distribution function (CDF) for the standard normal distribution.

STEP 13

The probability we want is the area to the right of the z-score in the standard normal distribution. This is given by 1−CDF(z)1 - \text{CDF}(z)1−CDF(z).P(xˉ>159)=1−CDF(−0.73)P(\bar{x} > 159) = 1 - \text{CDF}(-0.73)P(xˉ>159)=1−CDF(−0.73)

STEP 14

Using standard normal distribution tables or software, we find the CDF for z=−0.73z = -0.73z=−0.73.

STEP 15

The CDF for z=−0.73z = -0.73z=−0.73 is approximately 0.2327 (this value may vary slightly depending on the table or software used).

STEP 16

Now, calculate the probability that the sample mean is greater than 159 lb.P(xˉ>159)=1−0.2327P(\bar{x} > 159) = 1 - 0.2327P(xˉ>159)=1−0.2327

STEP 17

Calculate the final probability.P(xˉ>159)=1−0.2327≈0.7673P(\bar{x} > 159) = 1 - 0.2327 \approx 0.7673P(xˉ>159)=1−0.2327≈0.7673

STEP 18

The probability that the elevator is overloaded with 15 adult male passengers having a mean weight greater than 159 lb is approximately 0.7673, or 76.73%.

STEP 19

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An elevator has a capacity of 2385 lb for 15 passengers. Find the probability that 15 adult males with $\mu = 165$ lb, $\sigma = 32$ lb have a mean weight > 159 lb, indicating an overload. Does the elevator appear safe?

Now, plug in the given values for the maximum capacity and the number of passengers to calculate the maximum mean weight per passenger.
$\text{Maximum mean weight per passenger} = \frac{2385}{15}$

Calculate the maximum mean weight per passenger.
$\text{Maximum mean weight per passenger} = \frac{2385}{15} = 159 \text{ lb}$

First, we calculate the standard error of the mean (SEM), which is the standard deviation of the sampling distribution of the sample mean.
$SEM = \frac{\sigma}{\sqrt{n}}$
where $\sigma$ is the population standard deviation and $n$ is the sample size.

Now, plug in the given values for the population standard deviation and the sample size to calculate the SEM.
$SEM = \frac{32}{\sqrt{15}}$

Calculate the SEM.
$SEM = \frac{32}{\sqrt{15}} \approx 8.26 \text{ lb}$

Next, we calculate the z-score, which is the number of standard errors the sample mean is away from the population mean.
$z = \frac{\bar{x} - \mu}{SEM}$
where $\bar{x}$ is the sample mean, $\mu$ is the population mean, and SEM is the standard error of the mean.

Here, we are interested in the probability that the sample mean is greater than 159 lb. So, $\bar{x} = 159$ lb and $\mu = 165$ lb.
$z = \frac{159 - 165}{8.26}$

Calculate the z-score.
$z = \frac{159 - 165}{8.26} \approx -0.73$

The probability we want is the area to the right of the z-score in the standard normal distribution. This is given by $1 - \text{CDF}(z)$ .
$P(\bar{x} > 159) = 1 - \text{CDF}(-0.73)$

Using standard normal distribution tables or software, we find the CDF for $z = -0.73$ .

The CDF for $z = -0.73$ is approximately 0.2327 (this value may vary slightly depending on the table or software used).

Now, calculate the probability that the sample mean is greater than 159 lb.
$P(\bar{x} > 159) = 1 - 0.2327$

Calculate the final probability.
$P(\bar{x} > 159) = 1 - 0.2327 \approx 0.7673$