Stat 6.1
Find the indicated critical value. Z(.06)
Za denotes a Z score with an area to its right invnorm (.06,0,1) = 1.55
Find the area of the shaded region. The graph to the right depicts IQ scores of adults, and those scores are normally distributed with a mean of 100 and a standard deviation of 15.
normalcdf(95,120,100,15)=.5393
standard normal distribution
A normal distribution with a mean of 0 and a standard deviation of 1, and the total area under its density curve is 1.
Find the indicated area under the curve of the standard normal distribution; then convert it to a percentage and fill in the blank. About ______% of the area is between z=-2 and z=2 (or within 2 standard deviations of the mean).
95.45 %
Find the indicated area under the curve of the standard normal distribution; then convert it to a percentage and fill in the blank. About ______% of the area is between z=-3 and zequals3 (or within 3 standard deviations of the mean).
99.73% normalcdf(-3,3,0,1)
Assume that a randomly selected subject is given a bone density test. Bone density test scores are normally distributed with a mean of 0 and a standard deviation of 1. Draw a graph and find Upper P 5, the 5 th percentile. This is the bone density score separating the bottom 5 % from the top 95 %.
: p5 =z point where the area is shaded to the left invNorm(.05,0,1)= -1.64
Find the area of the shaded region. The graph depicts the standard normal distribution of bone density scores with mean 0 and standard deviation 1.
=.8554 Use DISTR button on calculator--> normalcdf(lower boundry, upper boundry, mean, stadnard deviation) Normalcdf(-1.06, 99999, 0, 1)
Find the indicated IQ score. The graph to the right depicts IQ scores of adults, and those scores are normally distributed with a mean of 100 and a standard deviation of 15.
Find the z score that corresponds to an area of 0.45. The z score is negative 0.13. Convert a standard normal random variable z to a normal random variable x by using the formula below, where mu is the mean and sigma is the standard deviation. x= u+(za) x=100+(-.13*15)= 98.1 In calc: invNorm(.65)=.39 .39*15= 5.85 5.85+100=105.8
Finding probabilities associated with distributions that are standard normal distributions is equivalent to _______.
Finding the area of the shaded region is equivalent to finding probabilities associated with distributions that are standard normal distributions because there is a correspondence between area and probability.
What does the notation z Subscript alpha indicate?
The expression z Subscript alpha denotes the z score with an area of alpha to its right.
Find the indicated IQ score. The graph to the right depicts IQ scores of adults, and those scores are normally distributed with a mean of 100 and a standard deviation of 15.
The indicated IQ score is 108 invNorm(.3204)=-.5 15*.5=7.5 8+100=108 Round after each term
What requirements are necessary for a normal probability distribution to be a standard normal probability distribution?
The mean and standard deviation have the values of u=0 and o=1.
The lengths of pregnancies are normally distributed with a mean of 269 days and a standard deviation of 15 days. a. Find the probability of a pregnancy lasting 308 days or longer. b. If the length of pregnancy is in the lowest 4%, then the baby is premature. Find the length that separates premature babies from those who are not premature.
The probability that a pregnancy will last 308 days or longer is z=(308-269)/15=2.6 normalcdf(2.6,9999,0,1) =.0047 Babies who are born on or before 243 days are considered premature. invNorm(.04)=-1.75 x=269+(-1.75*15)= 243
Finding standard normal distribution with the area
The total area under the curve is equal to 1. Use invnorm in calc with the percentage shown on the graph
For bone density scores that are normally distributed with a mean of 0 and a standard deviation of 1, find the percentage of scores that are a. significantly high (or at least 2 standard deviations above the mean). b. significantly low (or at least 2 standard deviations below the mean). c. not significant (or less than 2 standard deviations away from the mean).
a. The percentage of bone density scores that are significantly high is 2.28%. (Round to two decimal places as needed.) b. The percentage of bone density scores that are significantly low is 2.28%. (Round to two decimal places as needed.) c. The percentage of bone density scores that are not significant is 95.44%. (Round to two decimal places as needed.)
Find the indicated area under the curve of the standard normal distribution; then convert it to a percentage and fill in the blank. About ______% of the area is between z=-1 and z=1 (or within 1 standard deviation of the mean).
about 68.27 % of the area is between z=-1 and z=1 normalcdf(-1,1,0,1)
Find the indicated z score. The graph depicts the standard normal distribution of bone density scores with mean 0 and standard deviation 1.
invrNorm(.7823)= .78
Find the area of the shaded region. The graph depicts the standard normal distribution of bone density scores with mean 0 and standard deviation 1.
normalcdf(-.82,1.22,0,1) = .6827
Assume that a randomly selected subject is given a bone density test. Those test scores are normally distributed with a mean of 0 and a standard deviation of 1. Find the probability that a given score is less than negative 0.78 and draw a sketch of the region.
normalcdf(-99999,-.78,0,1)= .2177
Find the area of the shaded region. The graph to the right depicts IQ scores of adults, and those scores are normally distributed with a mean of 100 and a standard deviation of 15.
normalcdf(-99999,115,100,15)=.8413
Find the area of the shaded region. The graph to the right depicts IQ scores of adults, and those scores are normally distributed with a mean of 100 and a standard deviation of 15.
normalcdf(90, 99999, 100, 15) =.7475
Assume that an adult female is randomly selected. Suppose females have pulse rates that are normally distributed with a mean of 77.0 beats per minute and a standard deviation of 12.5 beats per minute. Find the probability of a pulse rate between 61 beats per minute and 73 beats per minute. (Hint: Draw a graph.)
x=61 and 73 u=77 o=12.5 (61-77)/12.5=-1.28 (73-77)/12.5=-.32 normcdf(-99999,-1.28,0,1) =.1003 normcdf(-99999,-.32,0,1) = .3745 .3745-.1003=.2742
Find the area of the shaded region. The graph to the right depicts IQ scores of adults, and those scores are normally distributed with a mean of 100 and a standard deviation of 15.
.2257
