Question

1. a final exam in math 160 has a mean of 73 with standard deviation 7.8. if 24 students are randomly selected, find the probability that the mean of their test scores is greater than 71. a) 0.9012 b) 0.8962 c) 0.5036 d) 0.0008 10) a final exam in math 160 has a mean of 73 with standard deviation 7.8. if 24 students are randomly selected, find the probability that the mean of their test scores is less than 70. a) 0.0278 b) 0.9699 c) 0.1006 d) 0.0301

Explanation:

Step1: Identify the formula for the standard error of the mean

The formula for the standard error of the mean $\sigma_{\bar{x}}=\frac{\sigma}{\sqrt{n}}$, where $\sigma$ is the population standard - deviation and $n$ is the sample size. Here, $\sigma = 7.8$ and $n = 24$, so $\sigma_{\bar{x}}=\frac{7.8}{\sqrt{24}}\approx1.59$.

Step2: Calculate the z - score for question 9

The z - score formula is $z=\frac{\bar{x}-\mu}{\sigma_{\bar{x}}}$, where $\bar{x}$ is the sample mean, $\mu$ is the population mean. For $\bar{x}=71$, $\mu = 73$, and $\sigma_{\bar{x}}\approx1.59$, we have $z=\frac{71 - 73}{1.59}\approx-1.26$. We want $P(\bar{X}>71)$, which is $P(Z>-1.26)=1 - P(Z\leq - 1.26)$. From the standard normal table, $P(Z\leq - 1.26)=0.1038$, so $P(Z>-1.26)=1 - 0.1038 = 0.8962$.

Step3: Calculate the z - score for question 10

For $\bar{x}=70$, $\mu = 73$, and $\sigma_{\bar{x}}\approx1.59$, we have $z=\frac{70 - 73}{1.59}\approx-1.89$. We want $P(\bar{X}<70)$, which is $P(Z<-1.89)$. From the standard normal table, $P(Z<-1.89)=0.0294\approx0.0301$.

Answer:

1. B. 0.8962
2. D. 0.0301