Math  /  Data & Statistics

QuestionP\square P  Technology  register if interested \frac{\text { Technology }}{\text { register if interested }} register if interested anajors, sample directly to working members of our ed your Google Drive already) \sim Dashboard I Rapialdontio - 8 Home Ebook-AP Statistics-Lavin-YF achieve bfwpub.com/courses/202med/ebook\#ebookshell Co ipst org bookmatiks - 2 - M 자 O ii (1) 图 目 6 - 目 \square E - ) 放 4 1. 1 - 6 - * 6 5 ล ! random variable XX be the number of cars (including SUVs and light trucks) they own. Here is the probability model if we ignore the few households that own more than 5 cars: \begin{tabular}{|l|c|c|c|c|c|c|} \hline Number of cars X: & 0 & 1 & 2 & 3 & 4 & 5 \\ \hline Probability: & 0.09 & 0.36 & 0.35 & 0.13 & 0.05 & 0.02 \\ \hline \end{tabular}
27. What's the expected number of cars in a randomly selected American household? a. 1.00 b. 1.75 c. 1.84 d. 2.00 e. 2.50
28. The standard deviation of XX is σY=1.08\sigma_{Y}=1.08. If many households were selected at random, which of the following would be the best interpretation of the value 1.08 ? a. The mean number of cars would be about 1.08 . Sign out Nov 13 4:50 us

Studdy Solution

STEP 1

What is this asking? What's the average number of cars owned by an American family, and what does the standard deviation of 1.08 mean? Watch out! Don't mix up the expected value (average) with the standard deviation (spread).

STEP 2

1. Calculate the Expected Value
2. Interpret the Standard Deviation

STEP 3

The *expected value*, also known as the *mean*, of a discrete random variable XX is calculated as: E(X)=xP(x) E(X) = \sum x \cdot P(x) Where xx represents each possible value of the variable, and P(x)P(x) is the probability of that value occurring.
Basically, it's a weighted average where each value is weighted by how likely it is to happen!

STEP 4

Let's apply this formula to our car problem.
We'll multiply each number of cars by its probability, and then add up all those products: E(X)=(00.09)+(10.36)+(20.35)+(30.13)+(40.05)+(50.02) E(X) = (0 \cdot 0.09) + (1 \cdot 0.36) + (2 \cdot 0.35) + (3 \cdot 0.13) + (4 \cdot 0.05) + (5 \cdot 0.02)

STEP 5

E(X)=0+0.36+0.70+0.39+0.20+0.10 E(X) = 0 + 0.36 + 0.70 + 0.39 + 0.20 + 0.10

STEP 6

E(X)=1.75 E(X) = 1.75 So, the expected number of cars in a randomly selected American household is **1.75**!

STEP 7

The **standard deviation** measures the typical spread or variability of the data around the mean.
A larger standard deviation means the data is more spread out, while a smaller standard deviation means the data is more clustered around the mean.

STEP 8

In this case, the standard deviation of XX is given as σX=1.08\sigma_X = 1.08.
This means that the number of cars owned by a randomly selected American household typically deviates from the mean (**1.75**) by about **1.08** cars.
So, we can expect a fair amount of variation in car ownership among households!

STEP 9

The expected number of cars per household is **1.75** (option b).
The standard deviation of 1.08 signifies the typical spread of car ownership around this average.

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