PNP transistor circuit

When switch $S$ is at position $1$:

$$V_B = R_2 \mathbin{/} (R_1 + R_2)*V_{CC} = 30 \mathbin{/} 130*-20=-4.62\mathrm{V}$$ $$R_{TH}=100*30 \mathbin{/} 100+30=23.1\mathrm{kOhm}$$ $$V_{TH}-I_ER_E-V_{BE}-I_BR_{TH}=0$$

where $I_B= I_E \mathbin{/} \beta$ and $V_{BE} =0.7\mathrm{V}$.

So,

$I_E= V_{TH}-V_{BE}\mathbin{/} (R_E + R_{TH} \mathbin{/} \beta)$

$I_E= (-4.62-0.7) \mathbin{/} (500+23100 \mathbin{/} 110)$

$I_E=-7.49\mathrm{mA}$

$$V_E = V_B + 0.7 = -4.62+0.7 = -3.92\mathrm{V}$$

$V_C = -V_{CC} + I_CR_C$

$V_C = -20 + 7.49*2000 \mathbin{/} 1000$

$$V_C = -5.02\mathrm{V}$$

$$V_{CE} = V_E -V_C = -5.02 - (-3.92) = -1.1\mathrm{V}$$

$$P = IV = -1.1 * -7.49 \mathbin{/} 1000 = 8.24\mathrm{mW}$$

Am I right?