1.) Suppose a certain species of fawns between 1 and 5 months old have a body we
ID: 3328393 • Letter: 1
Question
1.) Suppose a certain species of fawns between 1 and 5 months old have a body weight that is approximately normally distributed with mean µ = 22.4 kilograms and standard deviation = 4.9 kilograms. Let x be the weight of a fawn in kilograms. Convert the following x interval to a z interval. Round to the nearest hundredth. x < 36.1
A) z > –2.80 B) z < 2.80 C) z < 11.94 D) z < –2.80 E) z > 2.80
2.) 2. Let x = red blood cell (RBC) count in millions per cubic millimeter of whole blood. Suppose that for healthy females, x has an approximately normal distribution with mean µ = 4.7 and standard deviation = 0.3 Convert the following x interval from a laboratory test to a z interval. 4.0 < x < 5.5
A) 11.67 < z < 14.00 B) –4.00 < z < 14.00 C) –4.00 < z < –1.67 D) 11.67 < z < 29.67 E) 14.00 < z < 27.33
3.) Find the area under the standard normal curve over the interval specified below. To the left of z = 0
A) 0.999 B) 0.500 C) 0.023 D) 0.977 E) 0.159
4.) Find the area under the standard normal curve over the interval specified below. To the right of z =2
A) 0.159 B) 0.500 C) 0.841 D) 0.999 E) 0.023
5.) 5. Find the area under the standard normal curve over the interval specified below. Between z = –2 and 1
A) 0.819 B) 0.477 C) 0.341 D) 0.499 E) 0.136
6.) 6. Find the area under the standard normal curve over the interval specified below. Between z = 0.8 and z = 1.9
A) –0.759 B) 0.029 C) 0.471 D) 0.212 E) 0.183
7.) Let z be a random variable with a standard normal distribution. Find the indicated probability below. P(z –1.6)
A) 0.945 B) 0.055 C) 0.027 D) 0.973 E) 0.018
8.) Let z be a random variable with a standard normal distribution. Find the indicated probability below. P(z –0.9)
A) 0.816 B) 0.092 C) 0.158 D) 0.184 E) 0.316
10. Let z be a random variable with a standard normal distribution. Find the indicated probability below. P(–1.7 z –0.8)
A) 0.816 B) 0.167 C) 0.212 D) 0.788 E) 0.955
Explanation / Answer
(1) right choice is x < 36.1
z=(x-µ)/, for x=36.1, z=(36.1-22.4)/4.9=2.8
for x < 36.1 and z < 2.8
(2) non of the choice
z=(x-µ)/, for x=4, z=(4-4.7)/0.3=-2.33
for x=5, z=(5.5-4.7)/0.3=2.67
for 4<x<5.5 the z interval -2.33<z,2.67
(3) B.0.05
P(z<0)=0.5 ( using ms-excel =normsdist(0))
(4)E) 0.023
P(z>2)=1-P(z>2)=1-0.977=0.023
(5) A) 0.819
P(-2<z<1)=P(z<1)-P(z<-2)=0.842-0.023=0.818
(6)E) 0.183
P(0.8<z<1.9)=P(z<1.9)-P(z<0.8)=0.971-0.788=0.183
(7) B) 0.055
P(z<-1.6)=0.055 ( using ms-excel =normsdist(-1.6))
(8)A) 0.816
P(z –0.9)=1-P(z<-0.9)=1-0.184=0.816
(10) B) 0.167
P(–1.7 z –0.8)=P(z<-0.8)-P(z<-1.7)=0.212-0.045=0.167
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