Answer :
Final answer:
The probability that a student completes the activity in less than 45 seconds is approximately 0.8531, or 85.31%.
Explanation:
To find the probability that a student completes the activity in less than 45 seconds, we need to calculate the area under the normal distribution curve to the left of 45 seconds.
First, we need to standardize the value of 45 seconds using the formula:
Z = (X - μ) / σ
where Z is the standard score, X is the value we want to standardize, μ is the mean, and σ is the standard deviation.
Substituting the given values, we have:
Z = (45 - 39.1) / 5.6
Z ≈ 1.0536
Next, we need to find the probability corresponding to this standard score using a standard normal distribution table or a calculator. The probability can be interpreted as the area under the curve to the left of the standard score.
Looking up the value of 1.0536 in the standard normal distribution table, we find that the corresponding probability is approximately 0.8531.
Therefore, the probability that a student completes the activity in less than 45 seconds is approximately 0.8531, or 85.31%.
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To find the probability of completing a physical activity in less than a certain amount of time given a normal distribution, calculate the z-score using the formula (X - μ) / σ where X is the value, μ is the mean, and σ is the standard deviation and then find the associated probability from a z-table.
The student's question is asking about probability in a normal distribution. Let's say that the student wants to find the probability of completing the physical activity in less than 35 seconds. The first step is to calculate the 'z-score', which measures how many standard deviations away from the mean a certain value is. The formula for the z-score is: Z = (X - μ) / σ, where X is the value, μ is the mean, and σ is the standard deviation. Plugging the numbers in, we get: Z = (35 - 39.1) / 5.6 = -0.73. We would then consult a z-table to find the probability associated with this z-score. The z-table tells us that the probability of obtaining a z-score of -0.73 or less is approximately 0.23, so there is a 23% chance of completing the activity in less than 35 seconds.
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The amount of time to complete a physical activity in a PE class is approximately normally normally distributed with a mean of 39.1 seconds and a standard deviation of 5.6 seconds. a) What is the probability that a randomly chosen student completes the activity in less than 34.8 seconds? b) What is the probability that a randomly chosen student completes the activity in more than 45.3 seconds? c) What proportion of students take between 35.5 and 42.5 seconds to complete the activity? d) 75% of all students finish the activity in less than seconds. A leading magazine (like Barron's) reported at one time that the average number of weeks an individuat is unemployed is 34 weeks. Assume that for the population of all unemployed individuals the population mean length of unemployment is 34 weeks and that the population standard deviation is 7 weeks. Suppose you would like to select a random sample of 34 unemployed individuals for a follow-up study. Find the probabitity that a single randomly selected value is less than 35. P(x<35)= Find the probability that a sample of size n=34 is randomly selected with a mean less than 35. P(x<35)= Enter your answers as numbers accurate to 4 d simal places.
