Question

A sign in the elevator of a college library indicates a limit of 16 persons. In addition, there is a weight limit of 2,500 pounds. Assume that the average weight of students, faculty, and staff at this college is 150 pounds, that the standard deviation is 27 pounds, and that the distribution of weights of individuals on campus is approximately normal. A random sample of 16 persons from the campus will be selected. a. What is the mean of the sampling distribution of \(\bar{x}\) ? b. What is the standard deviation of the sampling distribution of \(\bar{x} ?\) c. What average weights for a sample of 16 people will result in the total weight exceeding the weight limit of 2,500 pounds? d. What is the probability that a random sample of 16 people will exceed the weight limit?

Short answer

a. The mean of the sampling distribution of \(\bar{x}\) is 150 pounds. \n b. The standard deviation of the sampling distribution of \(\bar{x}\) is 6.75 pounds. \n c. The average weights for a sample of 16 people that will result exceeding the total weight limit of 2,500 pounds is 156.25 pounds. \n d. The probability that a random sample of 16 people will exceed the weight limit is 0.1762.

Step-by-step solution

Calculation of the mean of the sampling distribution of \(\bar{x}\)

The mean of the sampling distribution of \(\bar{x}\) is equal to the population mean. Hence, it will be 150 pounds.

Calculation of the standard deviation of the sampling distribution of \(\bar{x}\)

The standard deviation of the sampling distribution of \(\bar{x}\), also known as standard error, is equal to the population standard deviation divided by the square root of the sample size. That is, \(\sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}} = \frac{27}{\sqrt{16}} = 6.75\) pounds.

Determination of the average weights for a sample of 16 people that will result in the total weight exceeding the elevator weight limit

If the total weight of 16 people exceeds 2,500 pounds, the average weight of one person that will exceed this limit is \(\bar{x}_E = \frac{2500}{16}= 156.25\) pounds.

Calculation of the probability a random sample of 16 people will exceed the weight limit

To find this probability, we need first to convert the average weight that will exceed the limit into z-score using the formula \(Z = \frac{\bar{x}_E - \mu}{\sigma_{\bar{x}}} = \frac{156.25 - 150}{6.75}= 0.93\). Checking this Z-value in the Z table, the corresponding probability value is 0.8238. However, since we need the probability of exceeding the weight limit, it will be \(P(x > Z) = 1 - P(x < Z) = 1 - 0.8238 = 0.1762\).

Key concepts

Normal Distribution

Understand the bell curve? That's what folks often picture when we chat about the normal distribution. It's math-talk for a pattern where most stuff—like people's weights—pile up around an average value, and the further you step away, the fewer things you find.

Picture lots of students on a college campus. If we plucked one at random, their weight would likely hover near that middle number, 150 pounds, mentioned in the exercise. Like rolling dice and expecting a bunch of 7s, with super high or super low weights being about as common as snake eyes or boxcars.

Standard Deviation

Think of standard deviation as the trusty ruler for measuring how wild numbers like weights or test scores are dancing around the average. In the elevator exercise, they gave us a standard deviation of 27 pounds. That means if we look at how spread out everyone's weight is on the scale, most are within 27 pounds of that typical 150-pound student.

It's like measuring how spiky or smooth our bell curve is. A teeny standard deviation? A super sleek hill. A big one? Think Rocky Mountains.

Probability

Chances are, you've heard about probability. It's all about gauging the odds, like betting on whether you'll pluck a red lollipop from a jar. In our elevator problem, probability tells us how likely it is to randomly assemble a group weighing more than the limit.

Imagine flipping a coin or playing 'Pin the Tail on the Donkey.' Will you score a bullseye, or will the pin end up in left field? That's what probability measures—how likely different outcomes are.

Z-score

Ever felt totally average or really out there? That's where z-score comes in. It's a number that tells us how your weight, height, or SAT score compares to the Joe Average at the center of the normal distribution.

For example, the exercise walks through turning an average weight that might break the elevator into a z-score. It’s like saying, 'Hey, you’re this many standard ruler-widths away from being Mr. or Ms. Average.' And when we work it out, we get a handy number we can map to a chance of something happening.