Question

A University of Maryland student is planning to build a new backyard flower planter upon returning home for Spring Break. The planter will be square in shape. Each side of the planter will consist of 50 bamboo sticks, stacked side by side. Let the diameter of a bamboo stick be a random variable, X. The supplier of the bamboo sticks states that the mean diameter of one bamboo stick is ux = 15mm, with a standard deviation of Ox= 1mm Part a Suppose the student randomly selects 50 bamboo sticks for building one side of the planter. Let X be the average diameter of these 50 bamboo sticks. In other words, X is a sample mean for a sample of 50 bamboo sticks. Recognizing that X is itself a random variable, compute its mean and standard deviation. Part b Refer to Problem (1), including Part (a). Suppose the width of one side of the planter equals the sum of the diameters of a random sample of 50 bamboo sticks. Compute a 95% confidence interval for the width of one side of the planter. In other words, compute a 95% confidence interval for 50X.

Answer 1

Part a:

Let X is a random variable shows the diameter if bamboo stick. Given that: 1 = 15, ở =1, n = 50 Since sample size is greater than equal to 30 so using CLT the sampling distribution of sample mean will be approximately normal with mean and SD as follows: Hz = l = 15 01 = = 1/sqrt( 50) = 0.1414

Part b:

Since sample size is greater than equal to 30 so using CLT the sampling distribution of sample sum will be approximately normal with mean and SD as follows: us = = 750 Os=o.Vn = 1/sqrt(50) = 7.0711

According to empirical rule, 95% data values lie within 2 standard deviation of mean. So requried interval is

us £ 205 = 750 +2.7.0711 = 750 – 14.1422 = (735.8578,764.1422)