Question

Test: Test 2 Ch 5-6 Time Remaining: 025309 Submit Test This 1 pt This Test: 30 pts does, fnd the mean and standard deviation 001650104902671 0 3400 OA. μ. (Round to one decimal place as needed) O B. The table is not a probability disaribution Fnd the standard deviation of the random variable x Select the comect choice below and, if necessary, ill in the answer box to complete your choice ○A.。“ (Rond to one decamal pace as needed ) O B. The table is not a probabiity distribution 丱@ DOLL

Answer 1

Solution:

Mean = $\mu$ = $\sum x.P(x)$

= 0*0.0165 + 1*0.1049 + 2*0.2671 + 3*3400 + 4*2164 + 5*0.0551

= 2.8002

   $\approx$ 2.8

Standard deviation = $\sigma$ = $\sqrt{\sum x^{2}p(x)-\mu ^{2}}$

Here,

$\sum x^{2}p(x)$ =  0²*0.0165 + 1²*0.1049 + 2²*0.2671 + 3²*3400 + 4²*2164 + 5²*0.0551

= 9.0732

$\sigma$ = $\sqrt{\sum x^{2}p(x)-\mu ^{2}}$ = $\sqrt{9.0732-2.8^{2}}$

Standard deviation = $\sigma$ =1.1