Question

Samples of aluminum-alloy channels were tested for stiffness. the following frequency distribution was obtained. The distribution is assumed to be normal.stiffness frequency2480 232440 352400 402360 332320 21a) What is the approximate mean of the population from which the sample were taken?b) What is the approximate standard deviation of the population from which the samples were taken?c) What is the approximate probability that stiffness would be less than 2350 for any given channel section?

Answer 1

The approximate mean and standard deviation of the population are calculated using weighted averages and the assumption of a normal distribution. The probability of a stiffness less than a specific value is determined using z-scores and standard normal distribution tables. The Central Limit Theorem supports the approximation of the distribution of sample means as normal.

Step-by-step explanation:

Approximate Mean and Standard Deviation of Population

To approximate the mean of the population from the sample provided, we need to calculate the weighted average of the stiffness frequencies. Then, assuming a normal distribution, we can estimate the standard deviation using the range and frequencies provided. The probability that the stiffness would be less than a certain value can be found using z-scores and the standard normal distribution table.

Approximate Probability Calculation

The probability that stiffness would be less than 2350 can be approximated by determining the z-score that corresponds to 2350 and then using the standard normal distribution table to find the probability that a value falls below this z-score.

Understanding Distributions and Calculations

The Central Limit Theorem indicates that the distribution of sample means will approximate a normal distribution, regardless of the shape of the population distribution, provided the sample size is sufficiently large. When analyzing distributions, we often use the sample mean and sample standard deviation to infer properties about the population, including making probability estimates for specific outcomes.

Answer 2

The answer is explained below

Step-by-step explanation:

stiffness frequency

2480 23

2440 35

2400 40

2360 33

2320 21

a) The mean for the population is calculated using the formula:

`mean (\mu)=(\Sigma f_ix_i)/(\Sigma f_i) \\=(x_1f_1+x_2f_2+.\ .\ .+x_nf_n)/(x_1+x_2+.\ .\ . +x_n) \\=((2480*23)+(2440*35)+(2400*40)+(2360*33)+(2320*21))/(23+35+40+33+21) =(365040)/(152)=2401.6`

b) The standard deviation is given by:

`\sigma=\sqrt{ (\Sigma f_i(x_i-\mu)^2)/(\Sigma f_1) } \\=\sqrt{ (f_1(x_1-\mu)^2+f_2(x_2-\mu)^2+.\ .\ .+f_n(x_n-\mu)^2)/(f_1+f_2+.\ .\ .+f_n) } \\=\sqrt{ (23(2480-2401.6)^2+35(2440-2401.6)^2+40(2400-2401.6)^2+33(2360-2401.6)^2+21(2320-2401.6)^2)/(23+35+40+33+21 )}\\=\sqrt{(390021.12)/(152) }= 50.7`

c) We have to find the z score for x = 2350. The z score is given by:

`z=(x-\mu)/(\sigma)=(2350-2401.6)/(50.7)=-1.02`

From the z table:

The probability that stiffness would be less than 2350 for any given channel section = P(x < 2350) = P(z < -1.02) = 0.1539 = 15.39%