Z-Scores and Standard Deviation

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Answer Key & Explanations

Z-Scores and Standard Deviation · Grade 10 · Worksheet 1

1. A normal distribution of test scores has a mean of 75 and a standard deviation of 8. If a student scores 91 on the test, what is their z-score? Answer: 2 Solution: To find the z-score for a test score of 91 in a normal distribution with mean 75 and standard deviation 8, we use the z-score formula. Write down the z-score formula.
   Full step-by-step solution

   To find the z-score for a test score of 91 in a normal distribution with mean 75 and standard deviation 8, we use the z-score formula. Step 1: Write down the z-score formula. The z-score formula is: z = (x - mean) / standard deviation Step 2: Identify the given values from the problem. x = 91 (the student's score) mean = 75 standard deviation = 8 Step 3: Substitute the given values into the formula. z = (91 - 75) / 8 Step 4: Perform the subtraction inside the parentheses. 91 - 75 = 16 So, z = 16 / 8 Step 5: Perform the division. 16 divided by 8 equals 2. So, z = 2 Step 6: Interpret the result. A z-score of 2 means the student's score is 2 standard deviations above the mean. Therefore, the z-score is 2.

2. Olivia is a marine biologist studying the lengths of adult sea otters in a protected coastal region. The lengths follow a normal distribution with a mean of 145 cm and a standard deviation of 7 cm. She measures one adult sea otter and finds its length to be 159 cm. What is the z-score for this otter's length, and what does it indicate about how this otter compares to the population? Answer: 2 Solution: Recall the z-score formula: z = (x - μ) / σ, where x is the individual value, μ is the mean, and σ is the standard deviation. Identify the values: x = 159 cm, μ = 145 cm, σ = 7 cm.
   Full step-by-step solution

   Step 1: Recall the z-score formula: z = (x - μ) / σ, where x is the individual value, μ is the mean, and σ is the standard deviation. Step 2: Identify the values: x = 159 cm, μ = 145 cm, σ = 7 cm. Step 3: Substitute into the formula: z = (159 - 145) / 7. Step 4: Calculate the numerator: 159 - 145 = 14. Step 5: Divide by the standard deviation: 14 / 7 = 2. Step 6: The z-score is 2. This means the otter's length is 2 standard deviations above the mean length of adult sea otters in the region. The answer is 2.

3. Kaia is analyzing the germination times of a rare native plant species. The germination times (in days) follow a normal distribution with a mean of 21 days and a standard deviation of 3 days. One particular seed from Kaia's experiment germinated in 15 days. Calculate the z-score for this germination time and interpret what this value means in terms of standard deviations from the mean. Answer: -2 Solution: Recall the z-score formula: z = (x - μ) / σ Identify the values from the problem: x = 15 days, μ = 21 days, σ = 3 days Substitute the values into the formula: z = (15 - 21) / 3 Calculate the numerator: 15 - 21 = -6 Divide by the standard deviation: -6 / 3 = -2 The z-score is -2, which means this…
   Full step-by-step solution

   Step 1: Recall the z-score formula: z = (x - μ) / σ Step 2: Identify the values from the problem: x = 15 days, μ = 21 days, σ = 3 days Step 3: Substitute the values into the formula: z = (15 - 21) / 3 Step 4: Calculate the numerator: 15 - 21 = -6 Step 5: Divide by the standard deviation: -6 / 3 = -2 Step 6: The z-score is -2, which means this germination time is 2 standard deviations below the mean. The answer is -2.

4. A normal distribution has μ = 82 and σ = 6. If x = 97, then z = ? Answer: 2.5 Solution: Recall the z-score formula: z = (x - μ) / σ Substitute the given values: z = (97 - 82) / 6 Calculate the numerator: 97 - 82 = 15 Divide by the standard deviation: 15 / 6 = 2.5 The z-score is 2.5 The answer is 2.5.
   Full step-by-step solution

   Step 1: Recall the z-score formula: z = (x - μ) / σ Step 2: Substitute the given values: z = (97 - 82) / 6 Step 3: Calculate the numerator: 97 - 82 = 15 Step 4: Divide by the standard deviation: 15 / 6 = 2.5 Step 5: The z-score is 2.5 The answer is 2.5.

5. A normal distribution of exam scores has a mean of 82 and a standard deviation of 6. A student's score is represented by a point 1.5 standard deviations above the mean on the bell curve. What is the student's z-score? Answer: 1.5 Solution: Recall the formula for z-score: z = (x - μ) / σ The problem states the student's score is 1.5 standard deviations above the mean This means (x - μ) / σ = 1.5 Therefore, the z-score is 1.5 The answer is 1.5.
   Full step-by-step solution

   Step 1: Recall the formula for z-score: z = (x - μ) / σ Step 2: The problem states the student's score is 1.5 standard deviations above the mean Step 3: This means (x - μ) / σ = 1.5 Step 4: Therefore, the z-score is 1.5 The answer is 1.5.

6. The average score on a national mathematics exam is 78 points with a standard deviation of 5 points. If Sarah scored 86 points on this exam, what is her z-score? Round your answer to two decimal places. Answer: 1.60 Solution: Recall the z-score formula. The z-score measures how many standard deviations a data point is from the mean. z = (x - μ) / σ x = individual score μ = population mean σ = population standard deviation Identify the given values from the problem.
   Full step-by-step solution

   Step 1: Recall the z-score formula. The z-score measures how many standard deviations a data point is from the mean. The formula is: z = (x - μ) / σ where: x = individual score μ = population mean σ = population standard deviation Step 2: Identify the given values from the problem. Mean (μ) = 78 Standard deviation (σ) = 5 Sarah’s score (x) = 86 Step 3: Substitute the values into the formula. z = (86 - 78) / 5 Step 4: Perform the subtraction in the numerator. 86 - 78 = 8 So, z = 8 / 5 Step 5: Divide 8 by 5. 8 / 5 = 1.6 Step 6: Round to two decimal places. 1.6 is already 1.60 when written to two decimal places. Step 7: Interpret the result. Sarah’s z-score is 1.60, meaning her score is 1.60 standard deviations above the mean. Final Answer: 1.60

7. A normal distribution has μ = 85 and σ = 9. What is the z-score for x = 112? Answer: 3.0 Solution: Recall the z-score formula: z = (x - μ) / σ Substitute the given values: z = (112 - 85) / 9 Calculate the numerator: 112 - 85 = 27 Divide by the standard deviation: 27 / 9 = 3.0 The z-score is 3.0, meaning the data point is 3.0 standard deviations above the mean.
   Full step-by-step solution

   Step 1: Recall the z-score formula: z = (x - μ) / σ Step 2: Substitute the given values: z = (112 - 85) / 9 Step 3: Calculate the numerator: 112 - 85 = 27 Step 4: Divide by the standard deviation: 27 / 9 = 3.0 Step 5: The z-score is 3.0, meaning the data point is 3.0 standard deviations above the mean. The answer is 3.0.

8. The heights of students in a Grade 10 class follow a normal distribution with a mean of 165 cm and a standard deviation of 6 cm. If a student is 178 cm tall, what is their z-score? Round your answer to two decimal places. Answer: 2.17 Solution: Mean (μ) = 165 cm Standard deviation (σ) = 6 cm Student's height (x) = 178 cm z = (x - μ) / σ z = (178 - 165) / 6 178 - 165 = 13 z = 13 / 6 13 ÷ 6 = 2.1666... 2.1666...
   Full step-by-step solution

   Step 1: Identify the given values Mean (μ) = 165 cm Standard deviation (σ) = 6 cm Student's height (x) = 178 cm Step 2: Write the z-score formula z = (x - μ) / σ Step 3: Substitute the values into the formula z = (178 - 165) / 6 Step 4: Calculate the numerator 178 - 165 = 13 Step 5: Divide by the standard deviation z = 13 / 6 Step 6: Calculate the division 13 ÷ 6 = 2.1666... Step 7: Round to two decimal places 2.1666... rounded to two decimal places is 2.17 The answer is 2.17.