Help Assignment 3 Part A All Students V4
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Grant MacEwan University *
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Course
151
Subject
Mathematics
Date
Feb 20, 2024
Type
docx
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4
Uploaded by JusticeBoulder1386
ASSIGNMENT 3 PART A FAQ HELP Finding a zscore:
Find the z score for z
0.01
Z
0.01 is the value of z such that the area to the right of it is 0.01. The area to the left of this z will have area 0.99 = 1 - 0.01 (area under a density curve is 1) 0.00
0.01
0.02
0.03
0.04
0.05
0.06
0.07
0.08
0.09
2.3
0.9893 0.9896 0.9898 0.9901 0.9904 0.9906 0.9909 0.9911 0.9913 0.9916 0.9901 is the area closest to 0.99 …. implies …
z
0.01
= 2.33
………………………..
Areas under a Density Curve:
Understand that the area under a density curve is 1. So P(X > #) = 1 – P(X <#)
And P(#
1
< X < #
2
) = 1 – P(X < #
1) – P(X > #
2) Finding the area above a given value for a N(mu, sigma)
Assume the wingspan of a monarch butterfly is normally distributed with a mean of 10.3 cm and a standard deviation of .6 cm. Find the probability that a monarch butterfly has a wingspan that exceeds 12 cm. P(X > 12)=P(Z > 2.83) = 1 – P(Z<2.83) = 1 – 0.9977 = 0.0023 Since z = (x - µ)/σ = (12-10.3)/0.6 = 2.83
Finding an area between two values for a N(mu, sigma)
Assume the wingspan of a monarch butterfly is normally distributed with a mean of 10.3 cm and a standard deviation of .6 cm. Find the probability that a monarch butterfly has a wingspan between 8.6 cm and 12 cm.
(8.6 < X < 12)=P(X < 12) – P(X < 8.6) = P(Z < 2.83) – P(X < -2.83)= 0.9977 – 0.0023 = 0.9954
as z = (x - µ)/σ
=(12 - 10.3)/0.6 = 2.83 for x =12 and z = (x - µ)/σ =
(8.6 - 10.3)/0.6 = -2.83 for x =8.6
Finding a percentile: The pth percentile of a data set is the number that divides the bottom p% of the data from the top (1-p)% of the data.
Clay targets follow a N(20 cm, 2 cm) distribution. What is the 20
th
percentile
for the targets? Want x such that P( X < x) = 0.20
In tables, we find that the closest area to 0.20 is 0.2005
It corresponds to a z value of -0.84
0.00
0.01
0.02
0.03
0.04
0.05
0.06
0.07
0.08
0.09
-0.8
0.2119 0.2090 0.2061 0.2033 0.2005 0.1977 0.1949 0.1922 0.1894 0.1867 z= (x - µ)/σ = (x – 20)/2 = -0.84, solve for x
x – 20 = (-0.84)(2) (multiplied both sides by 2)
x – 20 = -1.68 (add 20 to both sides), so x = -1.68 + 20
x = 18.32 cm Find the value above which some q% of the data under the curve lies.
Assume the wingspan of a monarch butterfly is normally distributed with a mean of 10.3 cm and a standard deviation of .6 cm. If a monarch butterfly with a wingspan in the top 1% of wingspans is considered “extraordinary”, find the wingspan that a monarch butterfly would need to have to be considered extraordinary. P(X > x) = P(Z > z) = 0.01
This is the same x and z corresponding to P(X<x) = P(Z<z) = 0.99 The z value corresponding to area of 0.99 in the left (bottom) tail is 2.33
Solve z = (x - µ)/σ
2.33 = (x – 10.3)/0.6 , 2.33(0.6) + 10.3 = 11.698 A butterfly with a wingspan of more than 11.698 cm
is “extraordinary”
Finding a “rejection” area below a given value for a N(mu, sigma)
A target is rejected by the quality control department if it is smaller than 15.9 cm. What percent of targets are
rejected? P(X < 15.9) = P(Z < (15.9-20)/2 ) = P(Z < -4.1/2) = P(Z <-2.05) = 0.0202
0.00
0.01
0.02
0.03
0.04
0.05
0.06
0.07
0.08
0.09
-2.0
0.0228 0.0222 0.0217 0.0212 0.0207 0.0202 0.0197 0.0192 0.0188 0.0183 Use the binomial distribution to find the probability of at least one “rejection” in n repeated independent trials, where the probability of “success” from trial to trial is the “rejection” probability.
Compute the probability of at least one
rejection among a random sample of 5 targets
. Probability reject a randomly picked target = p
=
P
(
X
<
15.9
)=
0.0202
. (from above)
Randomly pick five targets. Let Y
be the number of rejections. Assume Y
is BIN(n=5, p=0.0202)
P
(
Y ≥
1
)=
1
−
P
(
Y
=
0
)=
1
−
(
5
0
)
×
0.0202
0
¿
¿
1
−(
1
)(
1
)
¿
(used complement rule and binomial probability calculation)
*Understanding Parameters and Sample Estimates of a X
distribution (sampling distribution of sample means) *Assessing when the X
distribution is normal.
Important to realize:
Let a random sample of size n be taken from a population X with mean µ and standard deviation σ
RESULT 1: If X is N(µ, σ), then the sample mean X
of n independent observations has a N(µ,
σ
√
n
) distribution for all n.
RESULT 2: CENTRAL LIMIT THEOREM If X is any shape with mean mu and standard deviation σ, then the sample mean X
of n independent observations has a sampling distribution that in N( µ, σ
√
n
) for large n (generally n > 30)
*For a X
distributed as N(µ, σ
√
n
), find P(a
X
< #)
*For a X
distributed asN(µ , σ
√
n
) ), find P(a X
> #)
*For a X
distributed as N(µ , σ
√
n
), find probability a X
is within given number of units of the population mean
Example: Nightly hours of sleep are normally distributed with a mean of 8 hours and a standard deviation of 0.30 hours. *A random sample of n=36 young adults is of interest. What are the mean and standard deviation of the sampling
distribution of X
? What is the shape? Show formulas and explain why.
µ = 8 σ
√
n
= 0.3
√
36
= 0.3
6
= 0.05
shape is normal by the Central Limit Theorem as n > 30
*Find the probability the average hrs of sleep slept by a random sample of 36 young adults lies above 7.9 hrs. Find P(
X
> 7.9) = P(Z > -2) = 1 – P(Z < -2) = 1 – 0.0228 = 0.9972
using z = x
−
μ
σ
√
n
= 7.9
−
8
0.05
= -2 The probability that the average hours of sleep attained by a sample of n= 36 young adults lies above 7.9 hours is . 0.9972
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Example of “probability that X
is within certain number of units of the population mean μ
.
EXAMPLE: *A POPULATION has standard deviation of 0.3 .
If you take a random sample of size n=36, what is the probability that x
is within 0.1 units of the population mean μ
? We can solve this problem for any value of μ
.
X
∼
N
(
μ
X
=
μ,σ
X
=
σ
√
n
=
0.3
√
36
=
0.05
)
. by the Central Limit Theorem
We want P
(
μ
−
0.1
<
X
<
μ
+
0.1
)
, The z
-score for x
=
μ
−
0.1
and x
=
μ
+
0.1
are, respectively,
z
=
x
−
μ
X
σ
X
=
μ
−
0.1
−
μ
X
σ
X
=
μ
−
0.1
−
μ
σ
X
=
−
0.1
0.05
=−
2
and
z
=
x
−
μ
X
σ
X
=
μ
+
0.1
−
μ
X
σ
X
=
μ
+
0.1
−
μ
σ
X
=
0.1
0.05
=
2
Thus
P
(
μ
−
0.1
<
X
<
μ
+
0.1
)
=
P
(
−
2
<
Z
<
2
)
=
P
(
Z
<
2
)
−
P
(
Z
←
2
)
=
0.997
−
0.023
=
0.
974
Therefore, 97.4% of samples of size 36 have a X
value within 0.1 units of the population mean.
EXAMPLE: Suppose we had been given that μ
= 1 and asked “If you take a random sample of size n=36, what is
the probability that x
is within 0.1 units of the population mean μ
?” Then X
∼
N
(
μ
X
=
μ
=
1
,σ
X
=
σ
√
n
=
0.3
√
36
=
0.05
)
by the CLT
Then P
(
μ
−
0.1
<
X
<
μ
+
0.1
)
=
¿
P(1 - 0.1 <
X
<
1
+
0.1
) = Z scores are -0.9 and 1.01 are, respectively, z
¿
x
−
μ
X
σ
X
=
1
−
0.1
−
1
0.05
=
1
−
0.1
−
1
0.05
=
−
0.1
0.05
=−
2
and
z
=
x
−
μ
X
σ
X
=
1
+
0.1
−
1
0.05
=
1
+
0.1
−
1
0.05
=
0.1
0.05
=
2
P(-0.9 < X
<
1.01
¿
=
P
(
−
2
<
Z
<
2
)
=
P
(
Z
<
2
)
−
P
(
Z
←
2
)
=
0.997
−
0.023
=
0.
974