The weights of adobe bricks used for construction are...

The weights of adobe bricks used for construction are normally distributed with a mean of 3 pounds and a standard deviation of 0.25 pound. Assume that the weights of the bricks are independent and that a random sample of 20 bricks is selected. (a) What is the probability that all the bricks in the sample exceed 2.75 pounds? (b) What is the probability that the heaviest brick in the sample exceeds 3.75 pounds?

The weights of adobe bricks used for construction are normally distributed with a mean of 3 pounds and a standard deviation of 0.25 pound. Assume that the weights of the bricks are independent and that a random sample of 20 bricks is selected.
(a) What is the probability that all the bricks in the sample exceed 2.75 pounds?
(b) What is the probability that the heaviest brick in the sample exceeds 3.75 pounds?

Key Concepts

Independent Random Samples

When a sample is composed of independent random observations from a distribution, the joint probability of events concerning these observations can be calculated as the product of the individual probabilities. This is particularly useful when determining the likelihood that all sampled items meet a specified criterion, as it involves multiplying the probability for a single item raised to the power of the number of items.

Order Statistics

Order statistics involve the probabilities associated with the minimum or maximum values in a sample from a distribution. In this context, analysis of the maximum (or heaviest) brick's weight requires understanding the distribution of the largest observation in a sample, which is essential in determining extreme value probabilities.

Normal Distribution

The normal distribution is a continuous probability distribution characterized by its bell-shaped curve, described by its mean and standard deviation. It is fundamental in probability theory and statistics since many variables tend to follow this distribution under natural conditions. Its properties, including symmetry and the empirical rule, allow for straightforward probability calculations using standard techniques such as Z-scores.

Standardization

Standardization, or the Z-score transformation, is the process of converting a normal random variable to a standard normal variable by subtracting the mean and dividing by the standard deviation. This technique simplifies the computation of probabilities by reducing the problem to looking up standard values in the standard normal distribution table or using computational tools.

Transcript

00:02 In problem 27, we are going to calculate the probabilities of several random variables which are normally distributed.

00:10 Now, we are given that the weights of adobe bricks follows a normal distribution with mean 3 and standard deviation 0 .25.

00:20 And we have selected a random sample of 20 such bricks.

00:23 Now let's denote the weights of the 20 bricks by the random variables x1 x2 up to x20.

00:36 Now what we need to find is the probability that all the bricks weigh greater than 2 .75 pounds.

00:51 Now this is the same as probability x1 greater than 2 .75.

00:58 Comma x2 greater than 2 .75 and up to x20 greater than 2 .75.

01:11 Now note that it's given that the weights of the bricks are independent.

01:18 So we can rewrite this as probability x1 greater than 2 .75 times probability x2 greater than 2 .75 times up to probability x20 greater than 2 .75.

01:39 This is simply the property, i mean the definition of independent random variables.

01:46 And also note that it's given that these random variables are all identically distributed.

01:53 They all follow the normal distribution with mean 3 in standard deviation 0 .25.

01:58 So we can simplify this as probability x greater.

02:03 Than 2 .75 raise to the power 20 because there are 20 terms and all of them are basically the same probability.

02:13 Here x denotes any random variable that follows this normal distribution.

02:18 So we need to find this and this is equal to probability x minus 3 over 0 .25 greater than 2 .75 minus 3 over 0 .5 .5 and this becomes minus 0 .25 over 0 .25 equals minus 1.

02:46 And what we did here was convert x to a standard normal random variable.

02:53 So this is the same as probably z greater than minus 1, where z is a standard normal variable.

03:00 And of course, we have raised to the power 20 here...

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