-(q + 14) = -3 + (-14)
-q - 14 = -3 -14
-q - 14 = -11
-q = +14 - 11
-q= 3
1) To find the confidence interval
the sample mean x = 38 σ = 9; n = 85;
The confidence level is 95% (CL = 0.95) <span>CL = 0.95
so α = 1 – CL = 0.05
</span><span>α/2 = 0.025 </span>Z(α/2) = z0.025
The area to the right of Z0.025 is 0.025 and the area to the left of Z0.025 is 1 – 0.025 = 0.975
Z(α/2) = z0.025 = 1.645 This can be found using a computer, or using a probability table for the standard normal distribution.
<span>EBM = (1.645)*(9)/(85^0.5)=1.6058</span> x - EBM = 38 – 1.6058 = 36.3941 <span> x + EBM = 38 + 1.6058 = 39.6058
</span>The 95% confidence interval is (36.3941, 39.6058).
The answer is the letter D
<span>The value of 40.2 is <span>within the 95% confidence interval for the mean of the sample
</span></span>2) To find the confidence interval <span>
<span>the sample mean x = </span>76 σ = 20; n = 102; </span><span>
The confidence level is 95% (CL = 0.95) CL = 0.95
so α = 1 – CL = 0.05
α/2 = 0.025 Z(α/2) = z0.025
The area to the right of Z0.025 is 0.025 and the area to the
left of Z0.025 is 1 – 0.025 = 0.975
Z(α/2) = z0.025 = 1.645 This can be found using a computer,
or using a probability table for the standard normal distribution.
EBM = (1.645)*(20)/(102^0.5)=3.2575 x - EBM = 76 – 3.2575 = 72.7424 </span> x +
EBM = 76 + 3.2575 = 79.2575 <span>
The 95% confidence interval is (</span>72.7424 ,79.2575).<span>
The answer is the letter </span>A
and the letter D<span>
The value of 71.8 and 79.8 <span> are </span> outside<span>
the 95% confidence interval for the mean of the sample</span></span>
4 rock, 2 reggae, 2 country...total of 8
P(1st song is reggae) = 2/8 which reduces to 1/4
P(2nd song is reggae) = 1/7
P(both events happening) : 1/4 * 1/7 = 1/28
I think that would be C but idk tell me if im right
