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Math  /  Data & Statistics

Questionww-awualeks.com Win say 1 Wh Untitied document -- Traditional|Stats IN. Defonse I Stats INBA... Utah vs LA Stats \& P... Home - Northern Ess... Content (4) Kaicenat - Midterm Exam: Ch (4,2)2(1,2,3)3(1,2,3)4(1,2)5(1,2,3)C(1,24⁡(4,2) 2(1,2,3) 3(1,2,3) 4(1,2) 5(1,2,3) \operatorname{C(1,24} Cuestion 30 of 30 (1 point) I Question Attempt 1 of 1 Time Remaining: 50:2950: 29 Jonathan Español =30=30 < 19 20 21 22 23 F 25 =26=26 ≡27\equiv 27 ≡28\equiv 28 ≡29\equiv 29 =30=30
Put some air in your tires: Let XX represent the number of tires with low air pressure on a randomly chosen car. The probability distribution of XX is as follows. \begin{tabular}{c|ccccc} x\boldsymbol{x} & 0 & 1 & 2 & 3 & 4 \\ \hline P(x)\boldsymbol{P}(\boldsymbol{x}) & 0.1 & 0.3 & 0.3 & 0.2 & 0.1 \end{tabular}
Part: 0/20 / 2
Part 1 of 2 (a) Compute the mean μX\mu_{X}. Round the answer to three decimal places as needed. μX=\mu_{X}= □\square
Part: 1/21 / 2
Part 2 of 2 (b) Compute the standard deviation σX\sigma_{X}. Round the answer to three decimal places as needed. σX=□\sigma_{X}=\square

Studdy Solution

STEP 1

What is this asking? We're given the chances of a car having 0, 1, 2, 3, or 4 tires with low air pressure, and we need to find the average number of low-pressure tires and how spread out those numbers are. Watch out! Don't mix up the formulas for mean and standard deviation!
Also, remember that the standard deviation is the *square root* of the variance.

STEP 2

1. Calculate the mean.
2. Calculate the variance.
3. Calculate the standard deviation.

STEP 3

The **mean**, μX\mu_X, is the average number of low-pressure tires we'd expect to see on a random car.
To find it, we **multiply** each value of xx (the number of low-pressure tires) by its probability P(x)P(x), and then **add** all these products together.
This is like a weighted average, where the weights are the probabilities.

STEP 4

μX=0⋅0.1+1⋅0.3+2⋅0.3+3⋅0.2+4⋅0.1=0+0.3+0.6+0.6+0.4=1.9\begin{aligned} \mu_X &= 0 \cdot 0.1 + 1 \cdot 0.3 + 2 \cdot 0.3 + 3 \cdot 0.2 + 4 \cdot 0.1 \\ &= 0 + 0.3 + 0.6 + 0.6 + 0.4 \\ &= 1.9 \end{aligned} So, the **mean** number of low-pressure tires is μX=1.9\mu_X = 1.9.

STEP 5

The **variance**, σX2\sigma_X^2, measures how spread out the distribution is.
A larger variance means the number of low-pressure tires tends to be further from the mean.
To calculate it, we find the average of the *squared differences* between each value of xx and the mean.

STEP 6

[ \begin{aligned} \sigma_X^2 &= (0 - 1.9)^2 \cdot 0.1 + (1 - 1.9)^2 \cdot 0.3 + (2 - 1.9)^2 \cdot 0.3 + (3 - 1.9)^2 \cdot 0.2 + (4 - 1.9)^2 \cdot

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