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.
A z-score measures how many standard deviations an individual data point is from the mean of a population, assuming the population standard deviation is known and the sample size is large (typically n > 30). In contrast, a t-score is used when the sample size is small (n ≤ 30) or when the population standard deviation is unknown, relying on the sample's standard deviation instead. The t-distribution, which the t-score utilizes, is wider and has heavier tails than the normal distribution, reflecting more uncertainty in smaller samples. As sample sizes increase, the t-distribution approaches the normal distribution, making z-scores more applicable.
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.
If the Z-Score corresponds to the standard deviation, then the distribution is "normal", or Gaussian.
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.
The fundamental difference between the t statistic and a z score lies in the sample size and the underlying population variance. The t statistic is used when the sample size is small (typically n < 30) and the population variance is unknown, making it more appropriate for estimating the mean of a normally distributed population. In contrast, the z score is used when the sample size is large or when the population variance is known, as it assumes a normal distribution of the sample mean. Consequently, the t distribution is wider and has heavier tails than the z distribution, reflecting greater uncertainty in smaller samples.
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.
Sample score sheets for many beauty pageants can be viewed at the Related Link. Specific pageants may have sample score sheets on their websites.
Yes.