In: Math
Part 1
A company produces steel rods. The lengths of the steel rods are
normally distributed with a mean of 198.9-cm and a standard
deviation of 2.3-cm. For shipment, 22 steel rods are bundled
together.
Find P52, which is the average length
separating the smallest 52% bundles from the largest 48%
bundles.
P52 = _____ cm
Part 2
Scores for a common standardized college aptitude test are
normally distributed with a mean of 513 and a standard deviation of
96. Randomly selected men are given a Test Prepartion Course before
taking this test. Assume, for sake of argument, that the test has
no effect.
If 1 of the men is randomly selected, find the probability that his
score is at least 616.
P(X > 616) = __________
If 5 of the men are randomly selected, find the probability that
their mean score is at least 616.
P(M > 616) = __________
Part 3
A population of values has a normal distribution with μ=152.8
and σ=59.7. You intend to draw a random sample of size n=211.
Find the probability that a single randomly selected value is
between 140.9 and 164.3.
P(140.9 < X < 164.3) = __________
Find the probability that a sample of size n=211 is randomly
selected with a mean between 140.9 and 164.3.
P(140.9 < M < 164.3) = _____________
Part 4
A population of values has a normal distribution with μ=164.8
and σ=37.2. You intend to draw a random sample of size n=108.
Find the probability that a sample of size n=108 is randomly
selected with a mean between 154.1 and 169.8.
P(154.1 < M < 169.8) = _________
Part 5
A population of values has a normal distribution with μ=143.9
and σ=89.6. You intend to draw a random sample of size n=217.
Find the probability that a single randomly selected value is
between 133 and 162.1.
P(133 < X < 162.1) = ________
Find the probability that a sample of size n=217 is randomly
selected with a mean between 133 and 162.1.
P(133 < M < 162.1) = _______________
Part 6
A population of values has a normal distribution with μ=179 and
σ=98.2. You intend to draw a random sample of size n=75.
Find the probability that a sample of size n=75 is randomly
selected with a mean less than 181.3.
P(M < 181.3) = ___________
1)
µ = 198.9
σ = 2.3
n= 22
proportion= 0.5200
Z value at 0.52 =
0.050 (excel formula =NORMSINV(
0.52 ) )
z=(x-µ)/(σ/√n)
so, X=z * σ/√n +µ= 0.050 *
2.3 / √ 22 +
198.9 = 198.92(answer)
2)
µ = 513
σ = 96
n= 1
X = 616
Z = (X - µ )/(σ/√n) = ( 616
- 513 ) / ( 96 /
√ 1 ) = 1.073
P(X ≥ 616 ) = P(Z ≥
1.07 ) = P ( Z <
-1.073 ) = 0.1417
(answer)
------------------
µ = 513
σ = 96
n= 5
X = 616
Z = (X - µ )/(σ/√n) = ( 616
- 513 ) / ( 96 /
√ 5 ) = 2.399
P(X ≥ 616 ) = P(Z ≥
2.40 ) = P ( Z <
-2.399 ) = 0.0082
(answer)
3)
µ = 152.8
σ = 59.7
n= 1
we need to calculate probability for ,
140.9 ≤ X ≤ 164.3
X1 = 140.9 , X2 =
164.3
Z1 = (X1 - µ )/(σ/√n) = ( 140.9
- 152.8 ) / ( 59.7
/ √ 1 ) =
-0.20
Z2 = (X2 - µ )/(σ/√n) = ( 164.3
- 152.8 ) / ( 59.7
/ √ 1 ) =
0.19
P ( 140.9 < X <
164.3 ) = P (
-0.20 < Z < 0.19 )
= P ( Z < 0.19 ) - P ( Z
< -0.20 ) =
0.576 - 0.421 =
0.1554
(answer)
---------------------------------
µ = 152.8
σ = 59.7
n= 211
we need to calculate probability for ,
140.9 ≤ X ≤ 164.3
X1 = 140.9 , X2 =
164.3
Z1 = (X1 - µ )/(σ/√n) = ( 140.9
- 152.8 ) / ( 59.7
/ √ 211 ) =
-2.90
Z2 = (X2 - µ )/(σ/√n) = ( 164.3
- 152.8 ) / ( 59.7
/ √ 211 ) =
2.80
P ( 140.9 < X <
164.3 ) = P (
-2.90 < Z < 2.80 )
= P ( Z < 2.80 ) - P ( Z
< -2.90 ) =
0.997 - 0.002 =
0.9955
(answer)
4)
µ = 164.8
σ = 37.2
n= 108
we need to calculate probability for ,
154.1 ≤ X ≤ 169.8
X1 = 154.1 , X2 =
169.8
Z1 = (X1 - µ )/(σ/√n) = ( 154.1
- 164.8 ) / ( 37.2
/ √ 108 ) =
-2.99
Z2 = (X2 - µ )/(σ/√n) = ( 169.8
- 164.8 ) / ( 37.2
/ √ 108 ) =
1.40
P ( 154.1 < X <
169.8 ) = P (
-2.99 < Z < 1.40 )
= P ( Z < 1.40 ) - P ( Z
< -2.99 ) =
0.919 - 0.001 =
0.9174 (answer)
--------------
5)
µ = 143.9
σ = 89.6
n= 1
we need to calculate probability for ,
133 ≤ X ≤ 162.1
X1 = 133 , X2 =
162.1
Z1 = (X1 - µ )/(σ/√n) = ( 133
- 143.9 ) / ( 89.6
/ √ 1 ) =
-0.12
Z2 = (X2 - µ )/(σ/√n) = ( 162.1
- 143.9 ) / ( 89.6
/ √ 1 ) =
0.20
P ( 133 < X <
162.1 ) = P (
-0.12 < Z < 0.20 )
= P ( Z < 0.20 ) - P ( Z
< -0.12 ) =
0.580 - 0.452 =
0.1289 (answer)
--------------------
µ = 143.9
σ = 89.6
n= 217
we need to calculate probability for ,
133 ≤ X ≤ 162.1
X1 = 133 , X2 =
162.1
Z1 = (X1 - µ )/(σ/√n) = ( 133
- 143.9 ) / ( 89.6
/ √ 217 ) =
-1.79
Z2 = (X2 - µ )/(σ/√n) = ( 162.1
- 143.9 ) / ( 89.6
/ √ 217 ) =
2.99
P ( 133 < X <
162.1 ) = P (
-1.79 < Z < 2.99 )
= P ( Z < 2.99 ) - P ( Z
< -1.79 ) =
0.999 - 0.037 =
0.9621
(answer)
6)
µ = 179
σ = 98.2
n= 75
X = 181.3
Z = (X - µ )/(σ/√n) = ( 181.3
- 179.00 ) / (
98.200 / √ 75 ) =
0.20
P(X < 181.3 ) = P(Z ≤ 0.203
) = 0.5804
(answer)