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