Question

elevator has a placard stating that the maximum capacity is 3700 lb—26 passengers. so, 26 adult male passengers can have a mean weight of up to 3700/26 = 142 pounds. assume that weights of males are normally distributed with a mean of 190 lb and a standard deviation of 32 lb.
find the probability that 1 randomly selected adult male has a weight greater than 142 lb
find the probability that a sample of 26 randomly selected adult males has a mean weight greater than 142 lb
what do you conclude about the safety of this elevator?
(round to four decimal places as needed.)
c. does this elevator appear to be safe?
a. no, because there is a good chance that 26 randomly selected adult male passengers will exceed the elevator capacity.
b. yes, because 26 randomly selected adult male passengers will always be under the weight limit.
c. no, because 26 randomly selected people will never be under the weight limit.
d. yes, because there is a good chance that 26 randomly selected people will not exceed the elevator capacity

Explanation:

Step1: Calculate the z - score for a single adult male

The formula for the z - score is \(z=\frac{x-\mu}{\sigma}\), where \(x = 142\), \(\mu=190\), and \(\sigma = 32\).

$$z=\frac{142 - 190}{32}=\frac{- 48}{32}=-1.5$$

Step2: Find the probability for a single adult male

We want \(P(X>142)\). Using the standard normal distribution table, \(P(X>142)=1 - P(X\leq142)\).
Since \(P(X\leq142)\) corresponds to \(z=-1.5\), from the standard normal table \(P(Z\leq - 1.5)=0.0668\). So \(P(X>142)=1 - 0.0668 = 0.9332\)

Step3: Calculate the z - score for the sample mean

The formula for the z - score of the sample mean \(\bar{X}\) is \(z=\frac{\bar{x}-\mu}{\frac{\sigma}{\sqrt{n}}}\), where \(\bar{x}=142\), \(\mu = 190\), \(\sigma=32\), and \(n = 26\)

$$z=\frac{142-190}{\frac{32}{\sqrt{26}}}=\frac{-48}{\frac{32}{\sqrt{26}}}\approx\frac{-48}{6.27}\approx - 7.66$$

Step4: Find the probability for the sample mean

We want \(P(\bar{X}>142)\). Using the standard normal distribution table, \(P(\bar{X}>142)=1 - P(\bar{X}\leq142)\)
Since \(z\approx - 7.66\), \(P(Z\leq - 7.66)\approx0\). So \(P(\bar{X}>142)=1-0 = 1\) (approx, due to the extreme z - value, the probability is extremely close to 1)

Step5: Analyze the safety of the elevator

The probability that a sample of 26 adult males has a mean weight greater than 142 lb is extremely high (close to 1). This means there is a good chance that 26 randomly selected adult male passengers will exceed the elevator capacity.

Answer:

1. \(0.9332\)
2. Approximately \(1\)
3. A. No, because there is a good chance that 26 randomly selected adult male passengers will exceed the elevator capacity.