Answer:
7. μ=204.9 and σ=5.4968
8. μ=75.9 and σ=0.7136
9. p=0.9452
Step-by-step explanation:
7. - Given that the population mean =204.9 and the standard deviation is 81.90 and the sample size n=222.
-The sample mean,[tex]\mu_x[/tex]is calculated as:
[tex]\mu_x=\mu=204.9, \mu_x=sample \ mean[/tex]
-The standard deviation,[tex]\sigma_x[/tex] is calculated as:
[tex]\sigma_x=\frac{\sigma}{\sqrt{n}}\\\\=\frac{81.9}{\sqrt{222}}\\\\=5.4968[/tex]
8. For a random variable X.
-Given a X's population mean is 75.9, standard deviation is 9.6 and a sample size of 181
-#The sample mean,[tex]\mu_x[/tex] is calculated as:
[tex]\mu_x=\mu\\\\=75.9[/tex]
#The sample standard deviation is calculated as follows:
[tex]\sigma_x=\frac{\sigma}{\sqrt{n}}\\\\=\frac{9.6}{\sqrt{181}}\\\\=0.7136[/tex]
9. Given the population mean, μ=135.7 and σ=88 and n=59
#We calculate the sample mean;
[tex]\mu_x=\mu=135.7[/tex]
#Sample standard deviation:
[tex]\sigma_x=\frac{\sigma}{\sqrt{n}}\\\\=\frac{88}{\sqrt{59}}\\\\=11.4566[/tex]
#The sample size, n=59 is at least 30, so we apply Central Limit Theorem:
[tex]P(\bar X>117.4)=P(Z>\frac{117.4-\mu_{\bar x}}{\sigma_x})\\\\=P(Z>\frac{117.4-135.7}{11.4566})\\\\=P(Z>-1.5973)\\\\=1-0.05480 \\\\=0.9452[/tex]
Hence, the probability of a random sample's mean being greater than 117.4 is 0.9452
