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: Single cup probability** Calculate the z-score for $X=670$ mg: $$z = \frac{670 - 660}{35} = \frac{10}{35} = \frac{2}{7} \approx 0.2857$$ 4. **Find probability $P(X > 670)$:** This is the area to the right of $z=0.2857$ in the standard normal distribution. Using standard normal tables or a calculator: $$P(Z > 0.2857) = 1 - P(Z \leq 0.2857) \approx 1 - 0.6129 = 0.3871$$ So, the probability is approximately 0.3871. 5. **Part b: Sample mean probability** Sample size $n=10$, so the standard error is: $$\sigma_{\bar{x}} = \frac{35}{\sqrt{10}} = \frac{35}{3.1623} \approx 11.07$$ Calculate the z-score for sample mean $\bar{x} = 670$ mg: $$z = \frac{670 - 660}{11.07} = \frac{10}{11.07} \approx 0.9035$$ 6. **Find probability $P(\bar{X} > 670)$:** $$P(Z > 0.9035) = 1 - P(Z \leq 0.9035) \approx 1 - 0.8161 = 0.1839$$ So, the probability is approximately 0.1839. **Final answers:** - a) $P(X > 670) \approx 0.3871$ - b) $P(\bar{X} > 670) \approx 0.1839$