1. **Problem statement:** We have a normally distributed variable representing cholesterol content in a cup of ice cream with mean $\mu = 660$ mg and standard deviation $\sigma = 35$ mg. We want to find: a. The probability that a single cup has cholesterol content more than 670 mg. b. The probability that the mean cholesterol content of a sample of 10 cups is more than 670 mg. 2. **Formula and rules:** For a normal distribution, the standardized z-score is calculated by: $$ z = \frac{X - \mu}{\sigma} $$ For sample means, the standard deviation of the sample mean (standard error) is: $$ \sigma_{\bar{X}} = \frac{\sigma}{\sqrt{n}} $$ We use the z-score to find probabilities from the standard normal distribution. 3. **Part a: Probability for a single cup** Calculate the z-score for $X=670$ mg: $$ z = \frac{670 - 660}{35} = \frac{10}{35} = \frac{\cancel{10}}{\cancel{35}} = 0.2857 $$ Using standard normal tables or a calculator, find $P(X > 670) = P(Z > 0.2857)$. From the standard normal table, $P(Z < 0.2857) \approx 0.612$. Therefore, $$ P(Z > 0.2857) = 1 - 0.612 = 0.388 $$ So, the probability that a single cup has more than 670 mg cholesterol is approximately 0.388. 4. **Part b: Probability for sample mean of 10 cups** Calculate the standard error: $$ \sigma_{\bar{X}} = \frac{35}{\sqrt{10}} = \frac{35}{3.162} = 11.07 $$ Calculate the z-score for sample mean $\bar{X} = 670$: $$ z = \frac{670 - 660}{11.07} = \frac{10}{11.07} = 0.903 $$ Find $P(\bar{X} > 670) = P(Z > 0.903)$. From the standard normal table, $P(Z < 0.903) \approx 0.817$. Therefore, $$ P(Z > 0.903) = 1 - 0.817 = 0.183 $$ So, the probability that the sample mean of 10 cups is more than 670 mg is approximately 0.183. **Final answers:** - a. $P(X > 670) \approx 0.388$ - b. $P(\bar{X} > 670) \approx 0.183$