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If the Z-Score corresponds to the standard deviation, then the distribution is "normal", or Gaussian.
If n*p is greater than or equal to 5 AND n*q is greater than or equal to 5, you can use a normal distribution as an estimate for the binomial distribution. Recall, n is the number of trials, p is the probability of success of a trial, and q is 1-p.
Yes - but the distribution is not a normal distribution - this can happen with a distribution that has a very long tail.
The binomial distribution can be approximated with a normal distribution when np > 5 and np(1-p) > 5 where p is the proportion (probability) of success of an event and n is the total number of independent trials.
The normal distribution and the t-distribution are both symmetric bell-shaped continuous probability distribution functions. The t-distribution has heavier tails: the probability of observations further from the mean is greater than for the normal distribution. There are other differences in terms of when it is appropriate to use them. Finally, the standard normal distribution is a special case of a normal distribution such that the mean is 0 and the standard deviation is 1.
A half.
Scores on the SAT form a normal distribution with a mean of µ = 500 with σ = 100. What is the probability that a randomly selected college applicant will have a score greater than 640?
0.15542174161
0.419243340766
It is 0.017864
If the Z-Score corresponds to the standard deviation, then the distribution is "normal", or Gaussian.
There is no specific proportion: the answer depends on the level of significance beyond which subjects are considered to be outliers.
0.866
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If n*p is greater than or equal to 5 AND n*q is greater than or equal to 5, you can use a normal distribution as an estimate for the binomial distribution. Recall, n is the number of trials, p is the probability of success of a trial, and q is 1-p.
0.368 or 36.8%.And you should specify that it is a standard normal distribution.0.368 or 36.8%.And you should specify that it is a standard normal distribution.0.368 or 36.8%.And you should specify that it is a standard normal distribution.0.368 or 36.8%.And you should specify that it is a standard normal distribution.
Yes - but the distribution is not a normal distribution - this can happen with a distribution that has a very long tail.