z = 0.8416
No. By definition of the median, the median has 50 percent of the case below and 50 percent of the cases above. This has nothing to do with the cases being in a normal distribution.
In the normal distribution, the mean and median coincide, and 50% of the data are below the mean.
yes
34.1% of the data values fall between (mean-1sd) and the mean.
It would mean that the result was 2 standard deviations above the mean. Depending on the distribution of the variable, it may be possible to attach a probability to this, or more extreme, observations.It would mean that the result was 2 standard deviations above the mean. Depending on the distribution of the variable, it may be possible to attach a probability to this, or more extreme, observations.It would mean that the result was 2 standard deviations above the mean. Depending on the distribution of the variable, it may be possible to attach a probability to this, or more extreme, observations.It would mean that the result was 2 standard deviations above the mean. Depending on the distribution of the variable, it may be possible to attach a probability to this, or more extreme, observations.
In a normal distribution half (50%) of the distribution falls below (to the left of) the mean.
No. By definition of the median, the median has 50 percent of the case below and 50 percent of the cases above. This has nothing to do with the cases being in a normal distribution.
In the normal distribution, the mean and median coincide, and 50% of the data are below the mean.
The answer will depend on what the distribution is. Non-statisticians often assum that the variable that they are interested in follows the Standard Normal distribution. This assumption must be justified. If that is the case then the answer is 81.9%
A Z score of 300 is an extremely large number as the z scores very rarely fall above 4 or below -4. About 0 percent of the scores fall above a z score of 300.
yes
20 percent
2.576 sd
The answer will depend on the distribution of the weights. There is no basis for assuming that the distribution is normal, or even symmetrical.
150 feet
The basic formula we need is the Standard Normal Distribution formula (see related link) which is Z = (x - mean)/std dev. Solve for x, and we get: x = mean + z*std dev. From the Z table (see related link), area = .025 (your 2.5% in decimal form) we get the z value of -1.96. We can now solve for x. X = 400 + (-1.96)*50 or 302. Therefore, 302 is the point (value) below which 2.5% expenses fell (I assume it is $302).
In a standard distribution, the first quartile (Q1) represents the 25th percentile of the data. This means that 25% of the data falls below Q1, and consequently, 75% of the data falls above Q1. Therefore, 75% of the data is above Q1.