Question

Suppose we are testing a data set for normality. The data has a median of 80.25, a mean of 77.5, and a standard deviation of 2.1. Compute Pearson's Coefficient of Skewness for this data.

PC =
Suppose that the histogram looks roughly bell-shaped and there are no outliers. Can we say that this data is normally distributed?
A. Yes
B. No

Round your Pearson's Coefficient to at least three decimal places if applicable.

A laser printer is advertised to print text documents at a speed of 18 ppm (pages per minute). The manufacturer tells you that the printing speed is actually a Normal random variable with a mean of 17.62 ppm and a standard deviation of 4.5 ppm.

i) Find the mean and standard deviation of the sampling distribution of sample means with sample size n=34.

μX¯¯¯¯¯μX¯ =

σX¯¯¯¯¯σX¯ =

ii) Suppose that we draw a random sample of 34 printers. Find the probability that the mean of our sample will be greater than 18.62 ppm.

Probability =

iii) Suppose that we found all possible sample means for a sample of size 3434. Find the values that approximately 95% of the sample means would fall between.

Between __ and __ .

Give probabilities in four decimal places and all other answers to at least two decimal places where applicable.

Answer 1

Part 1)

i) Given : Median = 80.25 , mean = 77.5 and standard deviation = 2.1

Pearson's coefficient of skewness is

using emperical relation between mean, median and mode

PC = 3 * ( 77.5 - 80.25) / 2.1 = -3.929

PC = -3.929

Distribution is negatively skewed.

ii) If histogram looks bell-shaped and there is no outlier, then we can say that the distribution is normally distributed. The frequency curve of normal distribution is bell shaped.

Correct answer - Yes

Part 2) Consider the random variable

X : Printing speed of printer.

Given: X ~ N ( mu = 17.62, sigma= 4.5²)

i) n = number of printers randomly selected = 34

Xbar : Sample mean

Sampling distribution of mean is

ii) Required probability =

since

= P ( Z > 1.2953)

from normal probability table

P( Z > 1.2953) = 0.0976

P ( Mean of sample will be greater than 18.62 ppm) = 0.0976

iii) 95% confidence interval for mean is

Alpha: level of significance = 0.05

From normal probability table

Hence 95% confidence interval is

=( 16.107, 19.133)

95% of the sample means would fall between 16.107 and 19.133.