Ans : a) P( Z > 2.58) = 0.0049 ( 4 decimal places)
b) P ( Z < -1) = 0.1587 ( 4 decimal places)
c) P ( -1.5≦ Z < 5) = P ( -1.5 < Z < 5)
= (0.5- 0.0668) + ( 0.5 -0) = 0.9332 ( 4 decimal places)

2.Find the value of z if the area under a Standard Normal curve a)to the right of z is 0.3632;
b)to the left of z is 0.1131;
c)between 0 and z, with z > 0, is 0.4838;
d)between -z and z, with z > 0, is 0.9500.

Ans : a) z = + 0.35 ( find 0.5- 0.3632 = 0.1368 in the normal table) b) z = -1.21 ( find 0.5 – 0.1131 = 0.3869 in the normal table) c ) the area between 0 to z is 0.4838, z = 2.14
d) the area to the right of +z = ( 1-0.95)/2 = 0.025, therefore z = 1.96

3.Given the Normally distributed variable X with mean 18 and standard deviation 2.5, find a)P(X < 15);
b)the value of k such that P(X < k) = 0.2236;
c)the value of k such that P(X > k) = 0.1814;
d)P( 17 < X < 21).

Ans : X ~ N ( 18, 2.52)

a) P ( X < 15)
P ( Z < (15-18)/2.5) = P ( Z < -1.2) = 0.1151 ( 4 decimal places)

b) P ( X < k) = 0.2236
P ( Z < ( k – 18) / 2.5 ) = 0.2236
From normal table, 0.2236 = -0.76
(k-18)/2.5 = - 0.76, solve k = 16.1

c) P (X > k) = 0.1814
P ( Z > (k-18)/2.5 ) = 0.1814
From normal table, 0.1814 = 0.91
(k-18)/ 2.5 = 0.91, solve k = 20.275

d) P ( 17 < X < 21)
P ( (17 -18)/2.5 < Z < ( 21-18)/2.5)
P ( -0.4 < Z < 1.2) = 0.8849 – 0.3446 = 0.5403 ( 4 decimal places)

4.In a sample of 25 observations from a Normal Distribution with mean 98.6 and standard deviation 17.2, find:

Ans: a) n = 25, [pic] = ( = 98.6, [pic] = /n = 17.2/(25 = 3.44 [pic]( N...

...or variability of the data about the measurements of central tendency.
MEASUREMENTS OF CENTRAL TENDENCY The appropriateness of using the mean, median, or mode in data analysis is dependent upon the nature of the data set and its distribution (normal vs non-normal). The mean (denoted by x) is calculated by dividing the sum of the individual data points (where Σ equals “sum of”) by the number of observations (denoted by n). It is the arithmetic...

...skewed-right distribution with a mean of 10 minutes and a standard deviation of 8 minutes. Suppose 100 flights have been randomly sampled. Describe the sampling distribution of the mean waiting time between when the airplane taxis away from the terminal until the flight takes off for these 100 flights. a) Distribution is skewed-right with mean = 10 minutes and standard error = 0.8 minutes. b) Distribution is skewed-right with mean = 10...

...with an example
2
2
10
6
a) Describe the characteristics of Normal probability distribution.
b) In a sample of 120 workers in a factory, the mean and standard deviation of wages
were Rs. 11.35 and Rs.3.03 respectively. Find the percentage of workers getting wages
between Rs.9 and Rs.17 in the whole factory assuming that the wages are normally
distributed.
Characteristics of Normal probability distribution...

...X-bar Definition
1 x xi n i 1
Probability and statistics - Karol Flisikowski
n
Sampling Distribution of x-bar
How does x-bar behave? To study the behavior,
imagine taking many random samples of size n, and computing an x-bar for each of the samples. Then we plot this set of x-bars with a histogram.
Probability and statistics - Karol Flisikowski
Sampling Distribution of x-bar
Probability and statistics - Karol Flisikowski...

...Name: Ashley Lee
Class: HLT-362 Applied Statistics for Healthcare Professionals
Date: 04/01/2015
EXERCISE 18 • Mean, Standard Deviation, and 95% and 99% of the Normal Curve
1. Assuming that the distribution is normal for weight relative to the ideal and 99% of the male participants scored between (–53.68, 64.64), where did 95% of the values for weight relative to the ideal lie? Round your answer to two decimal places.
In order to find where 95%...

...at Weston Materials, Inc., a national manufacturer of unattached garages, reports that it takes two construction workers a mean of 32 hours and a standard deviation of 2 hours to erect the Red Barn model. Assume the assembly times follow the normaldistribution.
a. Determine the z values for 29 and 34 hours. What percent of the garages take between 32 hours and 34 hours to erect?
z(29) = (29-32)/2 = -3/2
z(34) = (34-32)/2 = 1
z(32) = 0
P(32 < x < 34) = P(0<...

...
Student Exploration: Sight vs. Sound Reactions
Vocabulary: histogram, mean, normaldistribution, range, standard deviation, stimulus
Prior Knowledge Questions (Do these BEFORE using the Gizmo.)
Most professional baseball pitchers can throw a fastball over 145 km/h (90 mph). This gives the batter less than half a second to read the pitch, decide whether to swing, and then try to hit the ball. No wonder hitting a baseball is considered one of the...

...points)
3. Suppose that a random sample of size 64 is to be selected from a population having [pic] and standard deviation 5.
(a) What are the mean and standard deviation of the [pic] sampling distribution? Can we say that the shape of the distribution is approximately normal? Why or why not? (10 points)
(b) What is the probability that [pic] will be within 0.5 of the population mean? (5 points)
(c) What is...