Probability of Cholesterol Content in Ice Cream Using Z-Scores

Math Problem Statement

Solution for #1.a:

Step1: Identify the parts of the problem.

Given: m = 46.2 𝑚𝑖𝑛𝑢𝑡𝑒𝑠; 𝜎 = 8 𝑚𝑖𝑛𝑢𝑡𝑒𝑠; 𝑋̅ = 43 𝑚𝑖𝑛𝑢𝑡𝑒𝑠; 𝑛 = 50 𝑠𝑡𝑢𝑑𝑒𝑛𝑡𝑠

Find: 𝑃(𝑋̅ < 43) Step 2: Use the formula to find the z-score.

𝑋̅ − m

𝒛 = ~~** s~~ =

√𝑛

43 − 46.2

√50

𝒛 = −𝟐. 𝟖𝟑 Step 3: Use the z-table to look up the z-score you calculated in step 2.

𝒛 = −2.83 has a corresponding area of 0.4977 **** Step 4: Draw a graph and plot the z-score and its corresponding area. Then, shade the part that you’re looking for: 𝑃(𝑋̅ < 43)

Since we are looking for the probability less than 43 minutes, the shaded part will be on the left part of – 2.83. Step 5: Subtract your z-score from 0.500.

𝑃(𝑋̅ < 43) = 0.500 − 0.4977

𝑃(𝑋̅ < 43) = 0.0023 Step 6: Convert the decimal in Step 5 to a percentage.

𝑃(𝑋̅ < 43) = 0.23% \ Therefore, the probability that a randomly selected 50 senior high school students will complete the examination in less than 43 minutes is 0.23%. No, it’s not reasonable since the probability is less than 1.

1. An electrical company claims that the average life of the bulbs it manufactures is 1 200 hours with a standard deviation of 250 hours. If a random sample of 100 bulbs is chosen, what is the probability that the sample mean will be between 1150 hours and 1 250 hours? Solution:

Step1: Identify the parts of the problem.

Given: m = 1200 ℎ𝑜𝑢𝑟𝑠; 𝜎 = 250 ℎ𝑜𝑢𝑟𝑠; 𝑛 = 100 𝑏𝑢𝑙𝑏𝑠

𝑋̅ = 1 150 & 1 250 ℎ𝑜𝑢𝑟𝑠

Unknown: 𝑃(1150 < 𝑋̅ < 1250) Step 2: Use the formula to find the z-score. 𝒛 =

𝑋̅ − m

~~** s~~ =

√𝑛

1150 − 1200

250

√100

𝒛 =

1250 − 1200

250

√100

𝒛 = −𝟐 𝒛 = 𝟐 Step 3: Use the z-table to look up the z-score you calculated in step 2.

𝒛 = ±2 has a corresponding area of 0.4772 **** Step 4: Draw a graph and plot the z-score and its corresponding area. Then, shade the part that you’re looking for: 𝑃(1150 < 𝑋̅ < 1250) Since we are looking for the probability between 1 150 hours and 1 250 hours, the shaded part will be between –2 and 2. Step 5: Add the two z-score values.

𝑃(1150 < 𝑋̅ < 1250) = 0.4772 + 0.4772

𝑃(1150 < 𝑋̅ < 1250) = 0.9544 Step 6: Convert the decimal in Step 5 to a percentage.

𝑃(1150 < 𝑋̅ < 1250) = 95.44% \ Therefore, the probability of randomly selected 100 bulbs to have a sample mean between 1 150 hours and 1 250 hours is 95.44%.

What’s More Let’s see how well you understood our discussion. At this point, I want you to solve the following problems. Show your complete solution by following the step-by- step procedure. 1. The average number of milligrams (mg) of cholesterol in a cup of a certain brand of ice cream is 660 mg, the standard deviation is 35 mg. Assume the variable is normally distributed.

a. If a cup of ice cream is selected, what is the probability that the cholesterol content will be more than 670 mg? b. If a sample of 10 cups of ice cream is selected, what is the probability that the mean of the sample will be larger than 670 mg?

Solution

Let's walk through the solution to the problems involving the cholesterol content of ice cream, step-by-step.

1. The average number of milligrams (mg) of cholesterol in a cup of a certain brand of ice cream is 660 mg, with a standard deviation of 35 mg. Assume the variable is normally distributed.

(a) If a cup of ice cream is selected, what is the probability that the cholesterol content will be more than 670 mg?

Step 1: Identify the parts of the problem.

Mean ( $\mu$ ) = 660 mg
Standard deviation ( $\sigma$ ) = 35 mg
Cholesterol content ( $X$ ) = 670 mg
We need to find $P(X > 670)$ .

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

The formula for the z-score is:

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

Substituting the values:

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

Step 3: Use the z-table to find the area corresponding to the z-score.

For $z = 0.29$ , the cumulative probability from the z-table is approximately 0.6141. This is the probability that the cholesterol content is less than 670 mg.

Step 4: Subtract the area from 1 to get the probability of more than 670 mg.

Since we are interested in $P(X > 670)$ , we subtract the cumulative probability from 1:

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

Step 5: Convert the decimal to a percentage.

$P(X > 670) = 0.3859 \times 100 = 38.59\%$

Therefore, the probability that a randomly selected cup of ice cream will have more than 670 mg of cholesterol is 38.59%.

(b) If a sample of 10 cups of ice cream is selected, what is the probability that the mean of the sample will be larger than 670 mg?

Step 1: Identify the parts of the problem.

Mean ( $\mu$ ) = 660 mg
Standard deviation ( $\sigma$ ) = 35 mg
Sample size ( $n$ ) = 10
Cholesterol content ( $X$ ) = 670 mg
We need to find $P(\bar{X} > 670)$ (the probability that the sample mean is greater than 670 mg).

Step 2: Calculate the standard error of the mean.

The formula for the standard error of the mean is:

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

Substitute the given values:

$SE = \frac{35}{\sqrt{10}} \approx \frac{35}{3.162} \approx 11.07$

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

For sample means, the z-score formula is:

[ z =

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Math Problem Analysis

Mathematical Concepts

Statistics
Probability
Normal Distribution
Z-Score

Formulas

Z-score formula: z = (X - μ) / σ
Standard error formula: SE = σ / √n
Cumulative probability formula using z-score

Theorems

Central Limit Theorem

Suitable Grade Level

Grades 11-12

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