Let's go through the problems step by step:

1.a. Find the probability that the cholesterol content of a randomly selected cup of ice cream will be more than 670 mg.

Step 1: Identify the parts of the problem.

• Mean cholesterol content $\mu = 660 \, \text{mg}$
• Standard deviation $\sigma = 35 \, \text{mg}$
• Cholesterol content threshold $X = 670 \, \text{mg}$

We are looking for $P(X > 670)$, which is the probability that the cholesterol content is greater than 670 mg.

Step 2: Use the formula to find the z-score.

The z-score formula is:

$z = \frac{X - \mu}{\sigma}$

Substitute the given values:

$z = \frac{670 - 660}{35} = \frac{10}{35} = 0.29$

Step 3: Use the z-table to find the probability.

Using a z-table, the area to the left of $z = 0.29$ is approximately 0.6141. This represents the probability that a randomly selected cup of ice cream has a cholesterol content less than 670 mg.

Since we need the probability of more than 670 mg:

$P(X > 670) = 1 - P(X < 670) = 1 - 0.6141 = 0.3859$

Step 4: Convert to percentage.

The probability that the cholesterol content will be more than 670 mg is:

$P(X > 670) = 0.3859 = 38.59\%$

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1.b. Find the probability that the mean cholesterol content of a sample of 10 cups will be larger than 670 mg.

Step 1: Identify the parts of the problem.

• Mean cholesterol content $\mu = 660 \, \text{mg}$
• Standard deviation $\sigma = 35 \, \text{mg}$
• Sample size $n = 10$
• Cholesterol content threshold $\bar{X} = 670 \, \text{mg}$

We are looking for $P(\bar{X} > 670)$, the probability that the mean cholesterol content of the sample is greater than 670 mg.

Step 2: Calculate the standard error.

The standard error (SE) of the mean is given by:

$SE = \frac{\sigma}{\sqrt{n}}$

Substitute the values:

$SE = \frac{35}{\sqrt{10}} = \frac{35}{3.162} \approx 11.08$

Step 3: Use the formula to find the z-score.

The z-score formula for the sample mean is:

$z = \frac{\bar{X} - \mu}{SE}$

Substitute the values:

$z = \frac{670 - 660}{11.08} = \frac{10}{11.08} \approx 0.90$

Step 4: Use the z-table to find the probability.

Using the z-table, the area to the left of $z = 0.90$ is approximately 0.8159. This represents the probability that the sample mean is less than 670 mg.

Since we are looking for the probability of a sample mean larger than 670 mg:

$P(\bar{X} > 670) = 1 - P(\bar{X} < 670) = 1 - 0.8159 = 0.1841$

Step 5: Convert to percentage.

The probability that the mean of the sample will be larger than 670 mg is:

$P(\bar{X} > 670) = 0.1841 = 18.41\%$

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Summary of Results:

• 1.a: The probability that the cholesterol content of a single cup is more than 670 mg is 38.59%.
• 1.b: The probability that the mean cholesterol content of a sample of 10 cups is more than 670 mg is 18.41%.

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Do you have any questions about the steps? Would you like more details on any part?

Related Questions:

1. If the standard deviation increases to 50 mg, what will the probability be in 1.a?
2. What is the probability that a cup has less than 640 mg of cholesterol?
3. How would the result change in 1.b if the sample size were 20 instead of 10?
4. If 5 cups of ice cream are selected, what is the probability that their mean cholesterol content will be less than 650 mg?
5. What is the z-score for a cholesterol content of 690 mg in a single cup of ice cream?

Tip:

Always check whether the problem is asking about a single observation or a sample mean, as this will determine whether to use the population standard deviation or the standard error of the mean.