Question

complete parts (a) and (b) below.
the number of dogs per household in a small town
dogs 0 1 2 3 4 5
probability 0.665 0.209 0.084 0.022 0.014 0.006

(a) find the mean, variance, and standard deviation of the probability distribution.
find the mean of the probability distribution.
μ = 0.5 (round to one decimal place as needed.)
find the variance of the probability distribution.
σ² = (round to one decimal place as needed.)

Explanation:

Step1: Recall Variance Formula

The formula for the variance \(\sigma^2\) of a discrete probability distribution is \(\sigma^2=\sum[(x - \mu)^2\cdot P(x)]\), where \(x\) is the value of the random variable, \(\mu\) is the mean, and \(P(x)\) is the probability of \(x\). We know \(\mu = 0.5\), \(x\) values are \(0,1,2,3,4,5\) and corresponding \(P(x)\) are \(0.665,0.209,0.084,0.022,0.014,0.006\).

Step2: Calculate Each \((x - \mu)^2\cdot P(x)\)

• For \(x = 0\): \((0 - 0.5)^2\cdot0.665=(- 0.5)^2\cdot0.665 = 0.25\cdot0.665 = 0.16625\)
• For \(x = 1\): \((1 - 0.5)^2\cdot0.209=(0.5)^2\cdot0.209 = 0.25\cdot0.209 = 0.05225\)
• For \(x = 2\): \((2 - 0.5)^2\cdot0.084=(1.5)^2\cdot0.084 = 2.25\cdot0.084 = 0.189\)
• For \(x = 3\): \((3 - 0.5)^2\cdot0.022=(2.5)^2\cdot0.022 = 6.25\cdot0.022 = 0.1375\)
• For \(x = 4\): \((4 - 0.5)^2\cdot0.014=(3.5)^2\cdot0.014 = 12.25\cdot0.014 = 0.1715\)
• For \(x = 5\): \((5 - 0.5)^2\cdot0.006=(4.5)^2\cdot0.006 = 20.25\cdot0.006 = 0.1215\)

Step3: Sum the Values

Sum all these values: \(0.16625+0.05225 + 0.189+0.1375+0.1715+0.1215\)
\(0.16625+0.05225=0.2185\); \(0.2185 + 0.189 = 0.4075\); \(0.4075+0.1375 = 0.545\); \(0.545+0.1715 = 0.7165\); \(0.7165+0.1215 = 0.838\)

Step4: Round to One Decimal Place

Rounding \(0.838\) to one decimal place gives \(0.8\) (since the second decimal is \(3\), which is less than \(5\), we round down? Wait, no, \(0.838\) to one decimal: look at the hundredth place, \(3\), so \(0.8\)? Wait, wait, maybe I made a miscalculation. Wait, let's recalculate the sum:

Wait, \(0.16625+0.05225 = 0.2185\); \(0.2185+0.189 = 0.4075\); \(0.4075 + 0.1375=0.545\); \(0.545+0.1715 = 0.7165\); \(0.7165+0.1215 = 0.838\). Yes, that's correct. Rounding to one decimal place: \(0.8\) (because the first decimal is \(8\), the next digit is \(3\), so we keep \(8\)). Wait, but let's check the formula again. Wait, maybe I used the wrong mean? Wait, the mean was given as \(0.5\). Wait, let's recalculate the mean to confirm. The mean formula is \(\mu=\sum[x\cdot P(x)]\). Let's recalculate \(\mu\):

\(0\cdot0.665 + 1\cdot0.209+2\cdot0.084 + 3\cdot0.022+4\cdot0.014+5\cdot0.006\)
\(=0 + 0.209+0.168+0.066+0.056+0.03\)
\(0.209+0.168 = 0.377\); \(0.377+0.066 = 0.443\); \(0.443+0.056 = 0.499\); \(0.499+0.03 = 0.529\approx0.5\) (rounded to one decimal). So \(\mu\approx0.5\) (more accurately \(0.529\)). Maybe we should use the more accurate \(\mu = 0.529\) for calculation. Let's recalculate \((x - \mu)^2\cdot P(x)\) with \(\mu = 0.529\):

• \(x = 0\): \((0 - 0.529)^2\cdot0.665=( - 0.529)^2\cdot0.665\approx0.279841\cdot0.665\approx0.1861\)
• \(x = 1\): \((1 - 0.529)^2\cdot0.209=(0.471)^2\cdot0.209\approx0.221841\cdot0.209\approx0.0464\)
• \(x = 2\): \((2 - 0.529)^2\cdot0.084=(1.471)^2\cdot0.084\approx2.163841\cdot0.084\approx0.1818\)
• \(x = 3\): \((3 - 0.529)^2\cdot0.022=(2.471)^2\cdot0.022\approx6.105841\cdot0.022\approx0.1343\)
• \(x = 4\): \((4 - 0.529)^2\cdot0.014=(3.471)^2\cdot0.014\approx12.047841\cdot0.014\approx0.1687\)
• \(x = 5\): \((5 - 0.529)^2\cdot0.006=(4.471)^2\cdot0.006\approx19.990841\cdot0.006\approx0.1199\)

Now sum these: \(0.1861+0.0464 = 0.2325\); \(0.2325+0.1818 = 0.4143\); \(0.4143+0.1343 = 0.5486\); \(0.5486+0.1687 = 0.7173\); \(0.7173+0.1199 = 0.8372\approx0.8\) (rounded to one decimal). So the variance is approximately \(0.8\) when rounded to one decimal place.

Answer:

\(0.8\)