Question

4.4 geometric distribution - 2026-sp-math-1401-co3 elementary statistics section co3 spring 2026 co, fall 2025 | wa consumer looking to buy a used red miata car will call dealerships until she finds a dealership that carries the car. she estimates the probability that any independent dealership will have the car will be 26%. we are interested in the number dealerships she must call.part (a)part (b)list the values that x may take on.○ x = 1, 2, 3, ..., 98, 99, 100○ x = 1, 2, 3, ..., 25, 26○ x = 1, 2, 3, ...○ x = 0, 1, 2, ...part (c)give the distribution of x.x ~ ( )part (d)on average, how many dealerships would we expect her to have to call until she finds one that has the car? (round your answer to the nearest whole number.)___ dealershipspart (e)find the probability that she must call at most 5 dealerships. (round your answer to four decimal places.)_part (f)find the probability that she must call 3 or 4 dealerships. (round your answer to four decimal places.)___

Explanation:

Part (b)

Step1: Identify X's possible values

X counts the number of dealerships called until success. It can be any positive integer (1, 2, 3, ...) since there's no upper limit to the number of calls needed.

Part (c)

Step1: Recognize distribution type

This is a geometric distribution, which models trials until the first success. The notation is $X \sim \text{Geometric}(p)$, where $p$ is the success probability.
$p = 0.26$, so $X \sim \text{Geometric}(0.26)$

Part (d)

Step1: Recall geometric mean formula

The expected value (mean) of a geometric distribution is $\mu = \frac{1}{p}$.
$\mu = \frac{1}{0.26}$

Step2: Calculate and round

$\frac{1}{0.26} \approx 3.846$, rounded to the nearest whole number is 4.

Part (e)

Step1: Define "at most 5" probability

$P(X \leq 5) = 1 - P(X > 5)$. For geometric distributions, $P(X > k) = (1-p)^k$.

Step2: Substitute values

$1-p = 1 - 0.26 = 0.74$, so $P(X \leq 5) = 1 - (0.74)^5$

Step3: Compute and round

$(0.74)^5 = 0.74 \times 0.74 \times 0.74 \times 0.74 \times 0.74 \approx 0.2219$
$1 - 0.2219 = 0.7781$

Part (f)

Step1: Define target probability

$P(X=3 \text{ or } X=4) = P(X=3) + P(X=4)$

Step2: Use geometric probability formula

The probability mass function is $P(X=k) = (1-p)^{k-1}p$.
For $k=3$: $P(X=3) = (0.74)^{2} \times 0.26$
For $k=4$: $P(X=4) = (0.74)^{3} \times 0.26$

Step3: Calculate each term

$(0.74)^2 = 0.5476$, so $P(X=3) = 0.5476 \times 0.26 \approx 0.1424$
$(0.74)^3 = 0.74 \times 0.5476 = 0.4052$, so $P(X=4) = 0.4052 \times 0.26 \approx 0.1053$

Step4: Sum and round

$0.1424 + 0.1053 = 0.2477$

Answer:

Part (b): $X = 1, 2, 3, ...$
Part (c): $X \sim \text{Geometric}(0.26)$
Part (d): 4 dealerships
Part (e): 0.7781
Part (f): 0.2477