the data set shows the time (in minutes) it took students to complete a quiz in class:
\{41, 46, 39, 44, 48, 36, 49, 28, 41, 40, 45, 37, 35, 41, 45\}. what is the probability that the students took less than or equal to 40 minutes to complete the quiz?

use the geogebra spreadsheet to find the mean and standard deviation. use these values to create a normal distribution for the data set. then use the geogebra probability calculator to solve the problem.

select the correct option below.

42.59%
54.71%
57.41%
49.52%

Question

the data set shows the time (in minutes) it took students to complete a quiz in class:
\{41, 46, 39, 44, 48, 36, 49, 28, 41, 40, 45, 37, 35, 41, 45\}. what is the probability that the students took less than or equal to 40 minutes to complete the quiz?

use the geogebra spreadsheet to find the mean and standard deviation. use these values to create a normal distribution for the data set. then use the geogebra probability calculator to solve the problem.

select the correct option below.

42.59%
54.71%
57.41%
49.52%

Accepted Answer

{ "quality": "clear", "question_count": 1, "skills_matched": [ "step_cot", "multiple_choice_formatter", "explain_then_answer", "quick" ], "needs_drawing": false, "concepts_used": [ "Sample Mean", "Sample Standard Deviation", "Normal Distribution Probability" ], "new_concepts": [], "current_concepts": [ "Sample Mean", "Sample Standard Deviation", "Normal Distribution Probability" ] } ## Calculate the sample mean $$ \begin{aligned} \text{Data set: } & \{41, 46, 39, 44, 48, 36, 49, 28, 41, 40, 45, 37, 35, 41, 45\} \ N = & 15 \ \sum x = & 41+46+39+44+48+36+49+28+41+40+45+37+35+41+45 = 625 \ \mu = & \frac{625}{15} \approx 41.6667 \end{aligned} $$ ## Calculate the sample standard deviation $$ \begin{aligned} \sum x^2 = & 41^2+46^2+39^2+44^2+48^2+36^2+49^2+28^2+41^2+40^2+45^2+37^2+35^2+41^2+45^2 \ = & 1681+2116+1521+1936+2304+1296+2401+784+1681+1600+2025+1369+1225+1681+2025 \ = & 26645 \ s = & \sqrt{\frac{\sum x^2 - \frac{(\sum x)^2}{N}}{N-1}} = \sqrt{\frac{26645 - \frac{625^2}{15}}{14}} = \sqrt{\frac{26645 - 26041.6667}{14}} = \sqrt{\frac{603.3333}{14}} \approx 6.5648 \end{aligned} $$ ## Calculate the normal probability $$ \begin{aligned} Z = & \frac{X - \mu}{s} = \frac{40 - 41.6667}{6.5648} \approx -0.2539 \ P(X \le 40) = & P(Z \le -0.2539) \approx 0.3998 \text{ (using population std dev } \sigma \approx 6.343 \text{ yields } Z \approx -0.2627 ightarrow 39.64\%) \end{aligned} $$ Let's recalculate using GeoGebra's standard sample standard deviation $s \approx 6.5648$: $$ P(X \le 40) \approx 39.98\% $$ If using population standard deviation $\sigma = \sqrt{\frac{603.3333}{15}} \approx 6.3421$: $$ Z = \frac{40 - 41.6667}{6.3421} \approx -0.2628 \implies P(Z \le -0.2628) \approx 39.64\% $$ Let's check the options: $42.59\%$, $54.71\%$, $57.41\%$, $49.52\%$. None of these match $40\%$. Let's re-verify the data values: $41, 46, 39, 44, 48, 36, 49, 28, 41, 40, 45, 37, 35, 41, 45$ Sum = 625. Mean = 41.6667. Let's check if one of the options matches a different calculation: If $\mu = 41.6667$ and $\sigma = 6.3421$: If $P(X \le 40)$ is asked: Let's check $Z$ for $42.59\%$: The Z-score for $42.59\%$ (0.4259) is $\approx -0.1868$. If $Z = -0.1868 = \frac{40 - \mu}{\sigma}$, then $\sigma = \frac{-1.6667}{-0.1868} \approx 8.92$. Let's check $Z$ for $49.52\%$: The Z-score for $0.4952$ is $\approx -0.012$. Let's check $Z$ for $42.59\%$ with sample standard deviation: If $s = 8.92$? No, $s \approx 6.56$. Wait, let's check if there is a typo in the dataset or if a value was read incorrectly. Let's calculate the mean and standard deviation of the dataset: Values: 41, 46, 39, 44, 48, 36, 49, 28, 41, 40, 45, 37, 35, 41, 45. Is it possible that the mean is different? Let's check the sum: 41+46 = 87 87+39 = 126 126+44 = 170 170+48 = 218 218+36 = 254 254+49 = 303 303+28 = 331 331+41 = 372 372+40 = 412 412+45 = 457 457+37 = 494 494+35 = 529 529+41 = 570 570+45 = 615. Wait! The sum is 615! Let's recalculate: 41+46+39+44+48+36+49+28+41+40+45+37+35+41+45 = 615. Yes! $615 / 15 = 41$. So the mean $\mu = 41$. Now let's recalculate $\sum x^2$: $41^2 = 1681$ $46^2 = 2116$ $39^2 = 1521$ $44^2 = 1936$ $48^2 = 2304$ $36^2 = 1296$ $49^2 = 2401$ $28^2 = 784$ $41^2 = 1681$ $40^2 = 1600$ $45^2 = 2025$ $37^2 = 1369$ $35^2 = 1225$ $41^2 = 1681$ $45^2 = 2025$ Sum of $x^2$ = 1681+2116+1521+1936+2304+1296+2401+784+1681+1600+2025+1369+1225+1681+2025 = 25645. Let's calculate sample variance: $$ s^2 = \frac{25645 - \frac{615^2}{15}}{14} = \frac{25645 - 25215}{14} = \frac{430}{14} \approx 30.7143 $$ $$ s = \sqrt{30.7143} \approx 5.5420 $$ Now calculate the Z-score for $X = 40$: $$ Z = \frac{40 - 41}{5.5420} \approx -0.1804 $$ Using the standard normal distribution: $$ P(Z \le -0.1804) \approx 0.4284 \approx 42.84\% $$ If using population standard deviation $\sigma = \sqrt{\frac{430}{15}} = \sqrt{28.6667} \approx 5.3541$: $$ Z = \frac{40 - 41}{5.3541} \approx -0.1868 $$ Using the standard normal distribution: $$ P(Z \le -0.1868) \approx 0.4259 = 42.59\% $$ This matches the option $42.59\%$ perfectly!

(A) 42.59%

(B) 54.71%

(C) 57.41%

(D) 49.52%

{ "subject": "Mathematics", "question_type": "Multiple Choice", "knowledge_point": [ "Mathematics", "Statistics", "Normal Distribution Probability" ] }

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QUESTION IMAGE

Question

Explanation:

Calculate the sample mean

Calculate the sample standard deviation

Calculate the normal probability

Answer: