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It approaches a normal distribution.
Because as the sample size increases the Student's t-distribution approaches the standard normal.
There is not enough information in the question to determine if the t-distribution is the appropriate model to use. If it is, then, with, a sample size of 95 the z-score for the Gaussian distribution is a suitable approximation. The probability is 0.199, approx.
If the Z-Score corresponds to the standard deviation, then the distribution is "normal", or Gaussian.
The estimated standard deviation goes down as the sample size increases. Also, the degrees of freedom increase and, as they increase, the t-distribution gets closer to the Normal distribution.
Z Score is (x-mu)/sigma. The Z-Score allows you to go to a standard normal distribution chart and to determine probabilities or numerical values.
z=(x-mean)/(standard deviation of population distribution/square root of sample size) T-score is for when you don't have pop. standard deviation and must use sample s.d. as a substitute. t=(x-mean)/(standard deviation of sampling distribution/square root of sample size)
The mean is 46.
Yes.
Sample score sheets for many beauty pageants can be viewed at the Related Link. Specific pageants may have sample score sheets on their websites.
It is impossible to determine the percentiles if you are given only the sample mean since percentiles are a measure of the spread of the data; the mean gives no information on that.
The standard normal distribution or the Gaussian distribution with mean 0 and variance 1.