Answer:
a) Mean=1.2
b) The standard deviation is
![\sigma=3.1](https://img.qammunity.org/2021/formulas/mathematics/college/3aat06dvhdtpthvdqtcc52g3rph80uq6nb.png)
Explanation:
Given that the discrete probability distribution below :
Outcome Probability
0 0.35
1 0.36
2 0.14
3 0.08
4 0.04
5 0.02
6 0.01
a) To find the mean of this distribution :
The formula is
![E(X)=\sum X.P(X)](https://img.qammunity.org/2021/formulas/mathematics/college/v9enkd0glps0zgvu7jhwbfudikfkmn6dma.png)
X P(X) XP(X)
![X^2P(X)](https://img.qammunity.org/2021/formulas/mathematics/high-school/ql60r6ft9pbr85yxhju8tqguw38hox405u.png)
0 0.35 0 0
1 0.36 0.36 0.36
2 0.14 0.28 1.12
3 0.08 0.24 2.16
4 0.04 0.16 2.56
5 0.02 0.1 2.5
6 0.01 0.06 2.16
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![\sum X^2P(X)=10.86](https://img.qammunity.org/2021/formulas/mathematics/college/m8au2wxnqo89r8vc6tuu2ra1xdjmzli7i6.png)
_________________________________________________
- Now substitute the value in the formula we get
![E(X)=\sum X.P(X)](https://img.qammunity.org/2021/formulas/mathematics/college/v9enkd0glps0zgvu7jhwbfudikfkmn6dma.png)
![E(X)=1.2](https://img.qammunity.org/2021/formulas/mathematics/college/mgshoazy31x4bfdvx0krzfufkjtde3g92d.png)
Therefore Mean=1.2
b) To find Standard Deviation :
The formula is
![\sigma=√(X^2P(X)-(XP(X))^2)](https://img.qammunity.org/2021/formulas/mathematics/college/zjg04z0s6il493cweolkcy678jluevc84i.png)
- Substitute the values i the formula we have
![\sigma=√(10.86-(1.2)^2)](https://img.qammunity.org/2021/formulas/mathematics/college/i2shm3rwloeja9ewlncm64ajumi5qyy258.png)
![=√(10.86-1.44)](https://img.qammunity.org/2021/formulas/mathematics/college/w3wp7qemz1gxvh5si0ilszs6fqwjw3u2i0.png)
![=√(9.42)](https://img.qammunity.org/2021/formulas/mathematics/college/s3s6sirt3lmrosrpch96ycky3oosldtp0s.png)
![=3.0692](https://img.qammunity.org/2021/formulas/mathematics/college/v5g38pyowc356ttu86o91l74jplr0z7yqe.png)
Therefore the standard deviation is
![\sigma=3.1](https://img.qammunity.org/2021/formulas/mathematics/college/3aat06dvhdtpthvdqtcc52g3rph80uq6nb.png)