chapter 5 review. find the area of the indicated region under the standard normal curve. use the...
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Find the indicated area under the standard normal curve. You may find the area using Table 4 or normalcdf. Sketch the standard normal curveTRANSCRIPT
NORMAL & STANDARD NORMAL DISTRIBUTIONS
Chapter 5 Review
Find the area of the indicated region under the standard normal curve. Use the table and show your work.
Find the areas to the left of the z-scores using Table 4 in the back of the book..Using Table 4, the area to the left of the z-score of
The area to the left of the z-score of
Subtract the two areas.
.8849β .5000=.3849
Find the indicated area under the standard normal curve. You may find the area using Table 4 or normalcdf.
To the left of .
Using Table 4, the area to the left of
β’ Using the graphing calculator:
β’ Set the mode to a float of 4.
β’ 2nd Vars (Distr)
β’ Normalcdf
Lower:Upper: 1.36
01
Paste to the home screen and push Enter
Sketch the standard normal curve
Find the indicated area under the standard normal curve. Use the table and show your work for credit.To the right of
The area at This is the area to the left of the z-score.
To find the area to the right of the z-score, subtract from 1.1.0000β .2578=.7422β’ Using the graphing calculator to check:
β’ 2nd Vars (Distr)
β’ Normalcdf
Lower:Upper: 1000
01
Paste to the home screen and push Enter
Sketch the standard normal curve
Find the indicated area under the standard normal curve. Use the table and show your work for credit.Between Sketch the standard normal curve
Find the areas to the left of the two z-scores using the table.
The area to the left of The area to the left of
Subtract the two areas. .5000β .0630=.4370β’ Using the graphing calculator to check:
β’ 2nd Vars (Distr)
β’ Normalcdf
Lower:Upper: 0
01
Paste to the home screen and push Enter
Find the indicated area under the standard normal curve. Use the table and show your work for credit.
To the left of
Find the area to the left of .1003
Find the area to the left of .8997
The area to the right of .1003
What is the clue word?
or
Which means weβre going to do what?
add
.1003+.1003=ΒΏ .2006
Find the probability of z occurring in the indicated region. Show your work for credit.
Use Table 4. The four-digit number in the table is also the probability.Find the areas to the left of both z-scores and subtract.
The area to the left of .3085
The area to the left of .8413
π (β0.5< π§<1.00 )=.8413β .3085
Find the indicated probability using the standard normal distribution. Show your work for credit.π (π§<1.45)
Use Table 4. The four-digit number in the table is also the probability.
The area to the left of .9265
π (π§<1.45 )=.9265
Find the indicated probability using the standard normal distribution. Show your work for credit.
π (π§>β0.95)
Use Table 4. The four-digit number in the table is also the probability.
The four digit number at .1711
is the area to the left. We must subtract this from 1.
π (π§>β0.95 )=1.0000β .1711=ΒΏ .8289
Find the indicated probability using the standard normal distribution.
π (β1.89<π§<0)
Sketch the curve.
Find the area to the left of the z-scores and subtract.
The area to the left of
The area to the left of
From table 4, the four-digit number represents both area and probability, so subtract the two areas.
π (β1.89<π§<0 )=.5000β .0294=.4706
.0294
.5000
Assume the random variable is normally distributed with a mean and standard deviation Find the indicated probability.
π (π₯<80)
Here we need to convert the x-value to a z-score in order to find the probability.
π§=π£πππ’πβππππ
π π‘ππππππ πππ£πππ‘πππ=π₯βππ
π§=80β865 =
β65 =β1.20
The area to the left of
π (π₯<80 )=π (π§<β1.20 )=.1151
.1151
Find the probability that the member selected at random is from the shaded area of the graph. Assume the variable x is normally distributed.
We need to convert the two x-values to z-scores.
π§=π£πππ’πβππππ
π π‘ππππππ πππ£πππ‘πππ=π₯βππ
π§=200β18635 .8 =
1435 .8=0.3911β0.39
Find the probability that the member selected at random is from the shaded area of the graph. Assume the variable x is normally distributed.
We need to convert the other x-value to a z-score.
π§=π£πππ’πβππππ
π π‘ππππππ πππ£πππ‘πππ=π₯βππ
π§=239β18635 .8 =
5335 .8=1.4804β1.48
Find the probability that the member selected at random is from the shaded area of the graph. Assume the variable x is normally distributed.
Using Table 4, find the probabilities to the two z-scores and subtract.The area to the left of
The area to the left of
Find the probability that the member selected at random is from the shaded area of the graph. Assume the variable x is normally distributed.
π (200<π₯<239 )=π (0.39<π§<1.48 )=.9306β .6517=.2789
We have been given x-values, a mean, and a standard deviation. The heights are normally distributed. So in order to find the probabilities, we need z-scores. And we will need to sketch three curves. Have fun!a) Find the probability that his height is less than 60 inches.
π (π₯<60 inches)
π§=π£πππ’πβππππ
π π‘ππππππ πππ£πππ‘πππ=π₯βππ
We need to find the z-score and go to Table 4.
The probability to the left of
π (π₯<60 inches )=π (π§<β1.20 )=.1151
π§=66β69.63.0 =
β3.63.0 =β1.20
.1151
We need to find two z-scores and go to Table 4.
b) Find the probability that his height is between 66 and 72 inches.
.π§=π£πππ’πβππππ
π π‘ππππππ πππ£πππ‘πππ=π₯βππ
π§=66β69.6
3 =β3.63 =β1.20
We already know one of the z-scores.
Now find the other z-score.
.
π§=π£πππ’πβππππ
π π‘ππππππ πππ£πππ‘πππ=π₯βππ
π§=72β69.63 .0 =
2.43 .0=0.80
The area to the left of
The area to the left of
Subtract the two areas (probabilities).
.,7881β .1151=.6730
π (66<π₯<72 )=π (β1.20<π§<0.80 )= .6730
Find the z-score that corresponds to the given cumulative area. If the area is not in the table, use the entry closest to the area.
0.7580
Go to Table 4 and look for the area.
Match it with the z-score.
.7580 can be found in the table.
The z-score which matches up with .7580 is 0.70
Find the z-score that corresponds to the given percentile.
π20
This is the 20th percentile, which converts to an area of .2000
Go to Table 4 and find the area nearest to .2000
.The area nearest to .2000 is .2005
The z-score that matches with the area is
π§=β0.84
Find the z-score that has 11.9% of the distributionβs area to its left.
Convert to a four-digit decimal. .1190
Go to Table 4 and find the area closes to
The area nearest to
The z-score which matches up to the area
π§=β1.18
Find the z-score that has 23.7% of the distributionβs area to its right. Convert to a four-digit decimal. .2370
This is the area to the right. We need to find the area to the left and find its z-score.
1.0000β .2370=ΒΏ .7630
Find the area nearest to .7642
The z-score for the area of
π§=0.72
a) What height represents the 95th percentile?
We are given the mean
We need to find the x-value. We also need to find the z-score.The 95th percentile converts to an area of .9500
The area in Table 4 closest to .9500 is .9505, which is a z-score of 1.65Use the following formula to find the value, x:
π₯=π+ π§π
π₯=64.1+(1.65)(2.71)
π₯=64.1+4.4715
π₯=68.5715 inches
π₯=68.6 inches
π₯=π+ π§π
b) What height represents the first quartile?
We are given the mean
We need to find the x-value. We also need to find the z-score.The first quartile converts to an area of .2500
The area in Table 4 closest to .2500 is .2514, which is a z-score of Use the following formula to find the value, x:
π₯=π+ π§π
π₯=64.1+(β0.67)(2.71)
π₯=64.1β1.8157
π₯=62.2843 inches
π₯=62.3 inches
π₯=π+ π§π