The heights of 1000 students are approximately normally distributed with a mean of 174.5 centimeters and a standard deviation of
6.9 centimeters. suppose 200 random samples of size 25 are drawn from this population and the means recorded to the nearest tenth of a centimeter. determine (a) the mean and standard deviation of the sampling distribution of x¯; (b) the number of sample means that falls between 172.5 and 175.8 centimeters inclusive; (c) the number of sample means falling below 172.0 centimeters.
The mean of a sampling distribution is approximately equal to the mean of the population. Given that the mean of the population is equal to 174.5, the mean of the sampling distribution is also this value.
The standard deviation of a sample distribution is equal to,
u(m) = u/sqrt n
Substituting the known values,
u(m) = 6.9 / sqrt 25 = 1.38
b. Get the z-score of both items,
z-score = (data point - mean) / standard deviation
z-score of 172.5 z-score = (172.5 - 174.5) / 1.38 = -1.49 This translates to 0.068.
z-score of 175.8 z-score = (175.8 - 174.5) / 1.38 = 0.94 This translates to 0.83.
The difference between the two z-scores is 0.762.
The number of samples with this height is 0.762(200) which is equal to approximately 152.
c. z-score of 172 centimeters
z-score = (172 - 174.5) / 1.38
z-score = -1.81 This translates to 0.03.
The number of people with this height from the sample is (0.03)(200) = 6
You start by writing your expression. Since each banner cost 28 per yard, you're going to multiply the price by the length by the number of banners she bought. (28)(17/8)(2)+(28)(25/8)=206.50
