Question

use the probability distribution to complete parts (a) and (b) below.
the number of school - related extracurricular activities per student
activities 0 1 2 3 4 5 6 7
probability 0.058 0.124 0.163 0.178 0.214 0.128 0.083 0.052
(a) find the mean, variance, and standard deviation of the probability distribution.
the mean is 3.3
(round to one decimal place as needed)
the variance is
(round to one decimal place as needed.)

Explanation:

Step1: Recall Variance Formula

For a discrete probability distribution, variance \(\sigma^2=\sum[(x - \mu)^2\cdot P(x)]\), where \(x\) is the random variable, \(\mu\) is the mean, and \(P(x)\) is the probability. Given \(\mu = 3.3\), \(x\) values: \(0,1,2,3,4,5,6,7\) and \(P(x)\) values: \(0.058,0.124,0.163,0.178,0.214,0.128,0.083,0.052\).

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

- For \(x = 0\): \((0 - 3.3)^2\cdot0.058=( - 3.3)^2\cdot0.058 = 10.89\cdot0.058\approx0.632\)
- For \(x = 1\): \((1 - 3.3)^2\cdot0.124=( - 2.3)^2\cdot0.124 = 5.29\cdot0.124\approx0.656\)
- For \(x = 2\): \((2 - 3.3)^2\cdot0.163=( - 1.3)^2\cdot0.163 = 1.69\cdot0.163\approx0.275\)
- For \(x = 3\): \((3 - 3.3)^2\cdot0.178=( - 0.3)^2\cdot0.178 = 0.09\cdot0.178\approx0.016\)
- For \(x = 4\): \((4 - 3.3)^2\cdot0.214=(0.7)^2\cdot0.214 = 0.49\cdot0.214\approx0.105\)
- For \(x = 5\): \((5 - 3.3)^2\cdot0.128=(1.7)^2\cdot0.128 = 2.89\cdot0.128\approx0.370\)
- For \(x = 6\): \((6 - 3.3)^2\cdot0.083=(2.7)^2\cdot0.083 = 7.29\cdot0.083\approx0.605\)
- For \(x = 7\): \((7 - 3.3)^2\cdot0.052=(3.7)^2\cdot0.052 = 13.69\cdot0.052\approx0.712\)

Step3: Sum all these values

Sum \(=0.632 + 0.656+0.275 + 0.016+0.105+0.370+0.605+0.712\)
\(=0.632+0.656 = 1.288\); \(1.288 + 0.275 = 1.563\); \(1.563+0.016 = 1.579\); \(1.579+0.105 = 1.684\); \(1.684+0.370 = 2.054\); \(2.054+0.605 = 2.659\); \(2.659+0.712 = 3.371\)

Step4: Round to one decimal place

Variance \(\approx3.4\) (Wait, wait, let's recalculate more accurately. Maybe my initial calculation had errors. Let's recalculate each term precisely:

- \(x=0\): \(3.3^2=10.89\), \(10.89\times0.058 = 10.89\times0.05 + 10.89\times0.008 = 0.5445+0.08712 = 0.63162\)
- \(x=1\): \(2.3^2 = 5.29\), \(5.29\times0.124 = 5.29\times0.1 + 5.29\times0.02 + 5.29\times0.004 = 0.529+0.1058+0.02116 = 0.65596\)
- \(x=2\): \(1.3^2 = 1.69\), \(1.69\times0.163 = 1.69\times0.1 + 1.69\times0.06 + 1.69\times0.003 = 0.169+0.1014+0.00507 = 0.27547\)
- \(x=3\): \(0.3^2 = 0.09\), \(0.09\times0.178 = 0.01602\)
- \(x=4\): \(0.7^2 = 0.49\), \(0.49\times0.214 = 0.10486\)
- \(x=5\): \(1.7^2 = 2.89\), \(2.89\times0.128 = 2.89\times0.1 + 2.89\times0.02 + 2.89\times0.008 = 0.289+0.0578+0.02312 = 0.36992\)
- \(x=6\): \(2.7^2 = 7.29\), \(7.29\times0.083 = 7.29\times0.08 + 7.29\times0.003 = 0.5832+0.02187 = 0.60507\)
- \(x=7\): \(3.7^2 = 13.69\), \(13.69\times0.052 = 13.69\times0.05 + 13.69\times0.002 = 0.6845+0.02738 = 0.71188\)

Now sum all these:

\(0.63162 + 0.65596 = 1.28758\)

\(1.28758 + 0.27547 = 1.56305\)

\(1.56305 + 0.01602 = 1.57907\)

\(1.57907 + 0.10486 = 1.68393\)

\(1.68393 + 0.36992 = 2.05385\)

\(2.05385 + 0.60507 = 2.65892\)

\(2.65892 + 0.71188 = 3.3708\)

Rounding \(3.3708\) to one decimal place: look at the second decimal, \(7\geq5\), so \(3.4\)? Wait, but maybe the mean was miscalculated? Wait the mean was given as 3.3. Let's check the mean calculation to confirm.

Mean \(\mu=\sum x\cdot P(x)\)

\(0\times0.058 + 1\times0.124 + 2\times0.163 + 3\times0.178 + 4\times0.214 + 5\times0.128 + 6\times0.083 + 7\times0.052\)

\(=0 + 0.124 + 0.326 + 0.534 + 0.856 + 0.64 + 0.498 + 0.364\)

Sum step by step:

\(0.124+0.326=0.45\); \(0.45+0.534=0.984\); \(0.984+0.856=1.84\); \(1.84+0.64=2.48\); \(2.48+0.498=2.978\); \(2.978+0.364=3.342\), which rounds to 3.3 (one decimal place). So mean is 3.342≈3.3.

Now variance is 3.3708≈3.4 when rounded to one decimal place? Wait, but let's check again. Wait, maybe I made a mistake in the formula. Variance is \(\sum x^2P(x)-\mu^2\). Let's try that formula to verify.

Calculate \(\sum x^2P(x)\):

\(0^2\times0.058 + 1^2\times0.124 + 2^2\times0.163 + 3^2\times0.178 + 4^2\times0.214 + 5^2\times0.128 + 6^2\times0.083 + 7^2\times0.052\)

\(=0 + 0.124 + 4\times0.163 + 9\times0.178 + 16\times0.214 + 25\times0.128 + 36\times0.083 + 49\times0.052\)

Calculate each term:

\(4\times0.163 = 0.652\)

\(9\times0.178 = 1.602\)

\(16\times0.214 = 3.424\)

\(25\times0.128 = 3.2\)

\(36\times0.083 = 2.988\)

\(49\times0.052 = 2.548\)

Now sum:

\(0.124 + 0.652 = 0.776\); \(0.776 + 1.602 = 2.378\); \(2.378 + 3.424 = 5.802\); \(5.802 + 3.2 = 9.002\); \(9.002 + 2.988 = 11.99\); \(11.99 + 2.548 = 14.538\)

Now \(\mu^2=(3.342)^2\approx11.17\) (Wait, no, \(\mu = 3.342\), so \(\mu^2 = 3.342\times3.342\approx11.17\))

Then variance \(=\sum x^2P(x)-\mu^2 = 14.538 - 11.17 = 3.368\approx3.4\) (when rounded to one decimal place). Wait, but the problem says "Round to one decimal place as needed". So 3.4? Wait, but let's check with the first method. The first method gave 3.3708, which is approximately 3.4. So the variance is approximately 3.4. Wait, but maybe I made a mistake in the first method. Wait, no, both methods should agree. Let's check \(\sum x^2P(x)=14.538\), \(\mu = 3.342\), \(\mu^2 = 3.342^2 = (3 + 0.342)^2 = 9 + 2\times3\times0.342 + 0.342^2 = 9 + 2.052 + 0.116964 = 11.168964\)

Then \(14.538 - 11.168964 = 3.369036\), which is approximately 3.4 when rounded to one decimal place. So the variance is 3.4.

Answer:

3.4