Question #177181

Find the area in each of the following. 

a. 𝑧 = −1.47 and 𝑧 = 1.64

b. 𝑧 = −0.44 and 𝑧 = 1.82

c. The region where 𝑧 > 1.92

d. The region where 𝑧 < 2.22

2. A biologist found the wingspans of a group of monary butterflies to be normally 

distributed with a mean of 52.2 mm and a standard deviation of 2.3 mm. What percent of 

the butterflies had a wingspan

a. less than 48.5 mm? 

b. between 50 and 55 mm?

3. A highway study of 8000 vehicles that passed by a checkpoint found that their speeds 

were normally distributed, with a mean of 61 mph and a standard deviation of 7 mph. 

a. How many of the vehicles had a speed of more than 68 mph? 

b. How many of the vehicles had a speed of less than 40 mph?




1
Expert's answer
2021-04-15T07:20:57-0400

1.

a.

P(1.47<Z<1.64)P(-1.47<Z<1.64)

=P(Z<1.64)P(Z1.47)=P(Z<1.64)-P(Z\leq-1.47)

0.9494970.070781=0.878716\approx0.949497-0.070781=0.878716

b.

P(0.44<Z<1.82)P(-0.44<Z<1.82)

=P(Z<1.82)P(Z0.44)=P(Z<1.82)-P(Z\leq-0.44)

0.9656200.329969=0.635661\approx0.965620-0.329969=0.635661

c.

P(Z>1.92)=1P(Z1.92)P(Z>1.92)=1-P(Z\leq1.92)

10.972571=0.027429\approx1-0.972571=0.027429

d.

P(Z<2.22)0.986791P(Z<2.22)\approx0.986791

10.972571=0.027429\approx1-0.972571=0.027429

2.

μ=52.2 mm,σ=2.3 mm\mu=52.2\ mm, \sigma=2.3\ mm

a.


P(X<48.5)=P(Z<48.552.22.3)P(X<48.5)=P(Z<\dfrac{48.5-52.2}{2.3})

P(Z<1.6087)0.0538\approx P(Z<-1.6087)\approx0.0538

5.38%5.38\%


b.


P(50<X<55)=P(X<55)P(X50)P(50<X<55)=P(X<55)-P(X\leq 50)

=P(Z<5552.22.3)(Z<5052.22.3)=P(Z<\dfrac{55-52.2}{2.3})-(Z<\dfrac{50-52.2}{2.3})

P(Z<1.2174)(Z<0.9565)\approx P(Z<1.2174)-(Z<-0.9565)




0.8882740.1694100.7189\approx 0.888274-0.169410\approx0.7189

71.89%71.89\%


3.

μ=61 mph,σ=7 mph\mu=61\ mph, \sigma=7\ mph

a.


P(X>68)=1P(X68)P(X>68)=1-P(X\leq 68)

=1P(Z68617)=1P(Z1)=1-P(Z\leq \dfrac{68-61}{7})=1-P(Z\leq 1)

10.841345=0.158655\approx 1-0.841345=0.158655

0.158655(8000)=12690.158655(8000)=1269


1269 vehicles


b.


P(X<40)=P(Z<40617)=P(Z<3)P(X<40)=P(Z<\dfrac{40-61}{7})=P(Z<-3)


0.001350\approx 0.001350

0.001350(8000)=110.001350(8000)=11


11 vehicles


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