Assuming a normal distribution:
If 16% are less than 307, then 84% are greater than 307.
Using single tailed normal tables we want the z value which gives 84%-50% = 34% = 0.34 to the left of the mean; this is z ≈ -0.995
z = (value - mean)/standard deviation
→ standard deviation = (value - mean)/z ≈ (307 - 312)/-0.995 ≈ 5.03
The z value of 322 is z ≈ (322 - 312)/5.03 ≈ 1.99
Using the single tailed tables give the value of the normal distribution for z = 1.99 as approx 0.477 to the right of the mean.
This means that 0.477 + 0.5 = 0.977 are less than 322
→ 1 - 0.977 = 0.023 = 2.3 % are greater than 322
→ probability greater than 322 is 2.3%
50 percent
5000 students participated in a certain test yielding a result that follows the normal distribution with means of 65 points and standard deviation of 10 points.(1) Find the probability of a certain student marking more than 75 points and less than 85 points inclusive.(2) A student needs more than what point to be positioned within top 5% of the participants in this test?(3) A student with more than what point can be positioned within top 100 students?i dont understand the question.. could you help me??pls....
If a student is picked at random what is the probability that he/she received an A on his/her fina?
It would be approximately normal with a mean of 2.02 dollars and a standard error of 3.00 dollars.
I think it depends on how many students are thinking of alphabet letters at the same time :) The first student will think of a letter from the English alphabet. Then, the second student will think of a letter. The probability that the second student will select what the first student selected is 1/26.
The probability is 0.5
Percent deviation is a measure of how much a value deviates, or differs, from a standard or expected value. It is calculated by taking the absolute difference between the measured value and the standard value, dividing by the standard value, and then multiplying by 100 to express it as a percentage.
The student's percent deviation is 25%, calculated as (|1.5 - 2.0| / 2.0) * 100%.
The probability is 10 percent.
The average IQ for a student in the UK is around 100, which is considered to be in the normal range. IQ scores are standardized to have a mean of 100 and a standard deviation of 15.
50 percent
The probability is indeterminate. I might ask a student or I might not.
Different examinations have different thresholds for A grades.
Context of this question is not clear because it is NOT a full question. However when attempting to estimate an parameter such as µ using sample data when the population standard deviation σ is unknown, we have to estimate the standard deviation of the population using a stastitic called s where _ Σ(x-x)² s = ▬▬▬▬ n -1 _ and estimator for µ , in particular x ........................................._ has a standard deviation of s(x)= s/√n and the statistic _ x - hypothesized µ T = ▬▬▬▬▬▬▬▬▬▬ s has a student's T distribution with n-1 degrees of freedom If n> 30 , then by the Central Limit Theorem, the T distribution approaches the shape and form of the normal(gaussian) probability distribution and the Z table may be used to find needed critical statistical values for hypothesis tests , p-values, and interval estimates.
If a student is picked at random what is the probability that he/she received an A on his/her fina?
5000 students participated in a certain test yielding a result that follows the normal distribution with means of 65 points and standard deviation of 10 points.(1) Find the probability of a certain student marking more than 75 points and less than 85 points inclusive.(2) A student needs more than what point to be positioned within top 5% of the participants in this test?(3) A student with more than what point can be positioned within top 100 students?i dont understand the question.. could you help me??pls....
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