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LESSON 2.6 • Normal Distributions: Finding Values from Areas 145

(b) Q 1 : (i) z = –0.67 gives an
deviation 1.1 inches. Helmets are mass-produced Exercises 15 and 16 refer to this setting. At fast-food
for all except the smallest 5% and the largest 5% restaurants, the lids for drink cups are made with a area to the left of about 0.25.
of head sizes. Soldiers in the smallest or largest small amount of flexibility, so they can be stretched −0.67 =− x = 3325.63
x 3668/511;
5% get custom-made helmets. What head sizes get across the mouth of the cup and then snugly secured.
custom-made helmets? When lids are too small or too large, customers can (ii) invNorm(area: 0.25,mean: 3668,
=
9. Long drive In 2019, the distribution of golfer Lexi get frustrated, especially if they end up spilling their SD: 511) 3323.3grams. Lesson 2.6
pg 143 Thompson’s driving distance had a mean of 276 drinks. At one restaurant, large drink cups require
23
yards. Assuming that her distribution of driving lids with a diameter of between 3.95 and 4.05 inches. Q 3 : (i) z = 0.67 gives an area
distance is approximately normal with an 80th The restaurant’s lid supplier claims that the diameter
percentile of 290 yards, calculate its standard of the large lids follows a normal distribution with to the left of about 0.75.
deviation. mean 3.98 inches and standard deviation 0.02 inch. 0.67 =− x = 4010.37
x 3668/511;
10. Get off your phone! According to a 2019 study Assume that the supplier’s claim is true. The supplier (ii) invNorm(area: 0.75,mean: 3668,
by RescueTime, people spend an average of 1.25 is considering two changes to reduce the percent of its
=
minutes on their smartphone each time they pick large-cup lids that are too small to 1%: (1) adjusting SD: 511) 4012.7grams. The first quartile
it up. If Noelle’s distribution of smartphone use the mean diameter of its lids, or (2) altering the pro- of the birth weight distribution
24
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is approximately normal with a mean of 1.25 min- duction process to decrease the standard deviation of
utes and a 97th percentile of 2 minutes, calculate its the lid diameters. is about 3323 grams and the
standard deviation. third quartile of the birth weight
15. Put a lid on the mean
11. Sub shop The lengths of footlong sub sandwiches distribution is about 4013 grams.
at a local sub shop follow an approximately (a) If the standard deviation remains at σ = 0.02 inch,
normal distribution with unknown mean µ and at what value should the supplier set the mean 15. (a) z = –2.33 gives an area to the
standard deviation 0.2 inch. If 20% of these diameter of its large-cup lids so that only 1% are
=
sandwiches are shorter than 11.7 inches, find the too small to fit? left of about 0.01: −2.33 3.95 − µ/0.02;
mean length µ. (b) What effect will the change in part (a) have on the µ = 3.9966 inches
percent of lids that are too large?
12. Rally time A tennis ball machine fires balls a dis-
tance that is approximately normally distributed. 16. Put a lid on the SD The supplier should set the mean
The mean distance, µ, is unknown and the stan- (a) If the mean diameter stays at µ = 3.98 inches, what diameter of its large-cup lids to 3.9966
dard deviation is 1.2 feet. If 5% of balls go farther value of the standard deviation will result in only
than 70 feet, find µ. 1% of lids that are too small to fit? inches. (b) By increasing the mean from
(b) What effect will the change in part (a) have on the 3.98 to 3.9966 inches the percentage of
Applying the Concepts percent of lids that are too large? lids that are too large will increase.
13. IQ test scores Scores on the Wechsler Adult Intel- 17. Quartiles Find the 25th percentile (Q 1 ) and the 75th 16. (a) z = –2.33 gives an area to the left
ligence Scale (a standard IQ test) for the 20-to-34 percentile (Q 3 ) of the standard normal distribution.
−
=
age group are approximately normally distributed 18. Deciles The deciles of any distribution are the val- of about 0.01: −2.33 3.95 3.98/SD;
with µ = 110 and σ = 25. ues at the 10th, 20th, . . . , 90th percentiles. The first SD = 0.0129 inch
(a) What percent of people aged 20 to 34 have IQs and last deciles are the 10th and the 90th percen-
between 125 and 150? tiles, respectively. What are the first and last deciles If the mean = 3.98 inches and SD =
(b) MENSA is an elite organization that admits as of the standard normal distribution? 0.0129 inch, 1% of lids will be too small.
members people who score in the top 2% on IQ 19. Flight times An airline flies the same route at the
tests. What score on the Wechsler Adult Intelligence same time each day. The flight time can be modeled (b) By decreasing the SD from 0.02 inch
Scale would an individual aged 20 to 34 have to by a normal distribution having unknown mean to 0.0129 inch, the percentage of too
earn to qualify for MENSA membership? and standard deviation. On 15% of days, the flight large lids will decrease.
14. Low birth weight Researchers in Norway ana- takes more than an hour. On 3% of days, the flight
lyzed data on the birth weights of 400,000 new- lasts 75 minutes or more. Use this information to 17. Qz = –0.67 gives an area to the left
: 1
borns over a 6-year period. The distribution of determine the mean and standard deviation of the of about 0.25; : 3 Qz = 0.67 gives an area
birth weight is approximately normal with a flight time distribution.
mean of 3668 grams and a standard deviation 20. Brush your teeth The amount of time Riccardo to the left of about 0.75.
25
of 511 grams. Babies who weigh less than spends brushing his teeth can be modeled by a
2500 grams at birth are classified as “low birth normal distribution with unknown mean and stan- 18. First decile: z =−1.28 gives an area
weight.” dard deviation. Riccardo spends less than 1 minute to the left of about 0.10; last decile:
(a) What percent of babies will be identified as low brushing his teeth about 40% of the time. He spends z =1.28 gives an area to the left of about
birth weight? more than 2 minutes brushing his teeth about 2%
(b) Find the first and third quartiles of the birth weight of the time. Use this information to determine the 0.90.
distribution. mean and standard deviation of this distribution.
19. We are looking for the value of z
with an area of 0.15 to the right and
0.85 to the left, and the value of z with
an area of 0.03 to the right and 0.97 to
the left. This is z =1.04 and z =1.88. We
11. z = –0.84 gives an area to the
07/09/20 1:58 PM
03_StarnesSPA4e_24432_ch02_088_153.indd 145 people aged 20 to 34 have IQs between 125 need to solve the following system of
left of about 0.2: −0.84 11.7 − µ/0.2; and 150. equations for µ and σ :1.0460 − µ σ
=
=
/
µ =11.87 inches (b) (i) z = 2.05 gives an area to the left of and 1.8875 − µ σ ; σ =17.86 minutes
=
/
x 110/25;
The mean length of footlong sub sandwiches about 0.98. 2.05 =− x =161.25 and µ = 41.43 minutes.
at this sub shop is 11.87 inches. (ii) invNorm(area: 0.98,mean:110, 20. We are looking for the value of
12. z =1.645 gives an area to the left of about SD:25) 161.34. An individual aged 20 to z with an area of 0.4 to the left and
=
0.95: 1.645 70 − µ/1.2; µ = 68.026 feet 34 would have to earn a score of 161.34 to the value of z with an area of 0.02 to
=
The mean distance that the tennis ball qualify for a MENSA membership. the right, and 0.98 to the left. This
machine fires a ball is 68.026 feet. 14. (a) (i) z = 2500 3668/511 =−2.29; the is z = –0.25 and z = 2.05. We need to
solve the following system of equations
−
=
13. (a) (i) z =125 110/250.60 and proportion of z-scores below z = –2.29 is for µ and σ −: 0.251 µ σ and
−
= −
/
−
z =150 110/251.6; the proportion of 0.0110. 2.05 =− µ σ; σ = 0.4348 minute and
=
/
2
z-scores between z = 0.60 and z =1.60 is µ =1.1087 minutes.
(ii) normalcdf(lower: –1000,upper: 2500,
0.9452–0.7257 = 0.2195. mean: 3668,SD: 511)0.0111. About 1.1% of
=
(ii) normalcdf(lower:125,upper:150, babies will be identified as low birth weight.
=
mean:110,SD: 25)0.2195. About 21.95% of
LESSON 2.6 • Normal Distributions: Finding Values from Areas 145
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