Question

Suppose 30% of all households in a certain town own three or more vehicles. what is the probability that a random sample of 70 households in that town will contain exactly 25 that have three or more vehicles? use a ti-83, ti-83 plus, or ti-84 calculator to find the probability.

Answer 1

The probability that a random sample of 70 households will contain exactly 25 that have three or more vehicles is approximately 0.1042.

Step-by-step explanation:

To calculate the probability using a TI-83, TI-83 Plus, or TI-84 calculator, we'll employ the binomial probability formula: P(X = k) = (n choose k) * p^k * (1 - p)^(n - k), where n is the sample size (70), k is the number of successes (25 households with three or more vehicles), and p is the probability of success in a single trial (30% or 0.30 in decimal form).

Using the calculator's functions, input the values into the formula: P(X = 25) = (70 nCr 25) * (0.30)^25 * (0.70)^(70 - 25). On the calculator, this would be computed as "70 nCr 25 * 0.30^25 * 0.70^45."

This computation results in the probability of exactly 25 out of 70 households having three or more vehicles. The answer, approximately 0.1042, signifies the likelihood of obtaining precisely 25 households with three or more vehicles in a random sample of 70.

Answer 2

The probability that a random sample of 70 households will contain exactly 25 that own three or more vehicles, given that 30% of all households in the town own three or more vehicles, is approximately 0.0585, or 5.85%.

To find the probability that exactly 25 out of 70 households own three or more vehicles when 30% of households in the town have this characteristic, we can use the binomial probability formula. However, since you've asked to use a TI-83, TI-83 Plus, or TI-84 calculator, the process involves using the binomial probability distribution function available on those calculators.

In a TI-84 calculator, this function is typically found under the "DISTR" menu. The specific function we would use is "binompdf" for binomial probability density function.

The syntax for the binompdf function is as follows:

```

binompdf(n, p, x)

```

where:

-
`\( n \)` is the number of trials (in this case, 70 households),

-
`\( p \)` is the probability of success on a single trial (30% or 0.3 for owning three or more vehicles),

-
`\( x \)` is the number of successes in
`\( n \)` trials (in this case, exactly 25 households).

Here's step-by-step how you would perform the calculation:

1. Press the ON button.

2. Press 2ND and then VARS to access the DISTR menu.

3. Scroll down to "A:binompdf(" and press ENTER.

4. Enter the number of trials
`\( n \)` which is 70, followed by a comma.

5. Enter the probability of success
`\( p \)` which is 0.3, followed by a comma.

6. Enter the number of successes
`\( x \)` which is 25.

7. Close the parenthesis and press ENTER.

The input on your calculator should look like this:

```

binompdf(70,0.3,25)

```

After pressing ENTER, the calculator will return the probability.

Since we are not using an actual calculator here, we can calculate this probability using Python which has functions to perform similar calculations as a TI calculator. Let's do that.

The probability that a random sample of 70 households will contain exactly 25 that own three or more vehicles, given that 30% of all households in the town own three or more vehicles, is approximately 0.0585, or 5.85%.