Answer:
[tex]-\frac{3(\sqrt{6}-\sqrt{2})}{2}\text{ or } \frac{-3\sqrt{6}+3\sqrt{2}}{2}}\text{ or }\frac{3(\sqrt{2}-\sqrt{6})}{2}[/tex]
Step-by-step explanation:
There are multiple ways to achieve and even express the exact answer to this problem. Because the exact value of [tex]6\cos(105^{\circ}})[/tex] is a non-terminating (never-ending) decimal, it does not have a finite number of digits. Therefore, you cannot express it as an exact value as a decimal, as you'd either have to round or truncate.
Solution 1 (Cosine Addition Identity):
Nonetheless, to find the exact value we must use trigonometry identities.
Identity used:
[tex]\cos(\alpha +\beta)=\cos \alpha \cos \beta-\sin \alpha \sin \beta[/tex]
Notice that [tex]45+60=105[/tex] and therefore we can easily solve this problem if we know values of [tex]\cos(45^{\circ})[/tex], [tex]\cos(60^{\circ})[/tex], [tex]\sin (45^{\circ})[/tex], and [tex]\sin(60^{\circ})[/tex], which is plausible as they are all key angles on the unit circle.
Recall from either memory or the unit circle that:
[tex]\cos(45^{\circ})=\sin(45^{\circ})=\frac{\sqrt{2}}{2}[/tex] [tex]\cos(60^{\circ})=\frac{1}{2}[/tex] [tex]\sin(60^{\circ})=\frac{\sqrt{3}}{2}[/tex]Therefore, we have:
[tex]\cos(105^{\circ})=\cos(45^{\circ}+60^{\circ}}),\\\cos(45^{\circ}+60^{\circ}})=\cos 45^{\circ}\cos 60^{\circ}-\sin 45^{\circ}\sin 60^{\circ},\\\cos(45^{\circ}+60^{\circ}})=\frac{\sqrt{2}}{2}\cdot \frac{1}{2}-\frac{\sqrt{2}}{2}\cdot \frac{\sqrt{3}}{2},\\\cos(105^{\circ})=\frac{\sqrt{2}}{4}-\frac{\sqrt{6}}{4},\\\cos(105^{\circ})={\frac{-\sqrt{6}+\sqrt{2}}{4}}[/tex]
Since we want the value of [tex]6\cos 105^{\circ}[/tex], simply multiply this by 6 to get your final answer:
[tex]6\cdot {\frac{-\sqrt{6}+\sqrt{2}}{4}}=\frac{-3\sqrt{6}+3\sqrt{2}}{2}}=\boxed{\frac{3(\sqrt{2}-\sqrt{6})}{2}}[/tex]
Solution 2 (Combination of trig. identities):
Although less plausible, you may have the following memorized:
[tex]\sin 15^{\circ}=\cos75^{\circ}=\frac{\sqrt{6}-\sqrt{2}}{4},\\\sin 75^{\circ}=\cos15^{\circ}=\frac{\sqrt{6}+\sqrt{2}}{4}[/tex]
If so, we can use the following trig. identity:
[tex]\cos(\theta)=\sin(90^{\circ}-\theta)[/tex] (the cosine of angle theta is equal to the sine of the supplement of angle theta - the converse is also true)
Therefore,
[tex]\cos (105^{\circ})=\sin (90^{\circ}-105^{\circ})=\sin(-15^{\circ})[/tex]
Recall another trig. identity:
[tex]\sin(-\theta)=-\sin (\theta)[/tex] and therefore:
[tex]\sin (-15^{\circ})=-\sin (15^{\circ})[/tex]
Multiply by 6 to get:
[tex]6\cos (105^{\circ})=-6\sin (15^{\circ})=-6\cdot \frac{\sqrt{6}-\sqrt{2}}{4}=\boxed{-\frac{3(\sqrt{6}-\sqrt{2})}{2}}[/tex] (alternative final answer).
If V= {i}, subset of V are?
Answer:
Defintion. A subset W of a vector space V is a subspace if
(1) W is non-empty
(2) For every v, ¯ w¯ ∈ W and a, b ∈ F, av¯ + bw¯ ∈ W.
Expressions like av¯ + bw¯, or more generally
X
k
i=1
aiv¯ + i
are called linear combinations. So a non-empty subset of V is a subspace if it is
closed under linear combinations. Much of today’s class will focus on properties of
subsets and subspaces detected by various conditions on linear combinations.
Theorem. If W is a subspace of V , then W is a vector space over F with operations
coming from those of V .
In particular, since all of those axioms are satisfied for V , then they are for W.
We only have to check closure!
Examples:
Defintion. Let F
n = {(a1, . . . , an)|ai ∈ F} with coordinate-wise addition and scalar
multiplication.
This gives us a few examples. Let W ⊂ F
n be those points which are zero except
in the first coordinate:
W = {(a, 0, . . . , 0)} ⊂ F
n
.
Then W is a subspace, since
a · (α, 0, . . . , 0) + b · (β, 0, . . . , 0) = (aα + bβ, 0, . . . , 0) ∈ W.
If F = R, then W0 = {(a1, . . . , an)|ai ≥ 0} is not a subspace. It’s closed under
addition, but not scalar multiplication.
We have a number of ways to build new subspaces from old.
Proposition. If Wi for i ∈ I is a collection of subspaces of V , then
W =
\
i∈I
Wi = {w¯ ∈ V |w¯ ∈ Wi∀i ∈ I}
is a subspace.
Proof. Let ¯v, w¯ ∈ W. Then for all i ∈ I, ¯v, w¯ ∈ Wi
, by definition. Since each Wi
is
a subspace, we then learn that for all a, b ∈ F,
av¯ + bw¯ ∈ Wi
,
and hence av¯ + bw¯ ∈ W. ¤
Thought question: Why is this never empty?
The union is a little trickier.
Proposition. W1 ∪ W2 is a subspace iff W1 ⊂ W2 or W2 ⊂ W1.
i hope this helped have a nice day/night :)
according to byu idaho enrollment statisct there are 1200 femaile studnet here on campus during any given semester of those 3500 have serced a msion what is the probability that a radnoly selcted femal studne ton cmapus wil have served a mission g
Answer:
0.2917 = 29.17% probability that a randomly selected female student on campus will have served a mission.
Step-by-step explanation:
A probability is the number of desired outcomes divided by the number of total outcomes.
In this question:
1200 female students, out of them, 350 have served a mission. So
[tex]p = \frac{350}{1200} = 0.2917[/tex]
0.2917 = 29.17% probability that a randomly selected female student on campus will have served a mission.
A wooden log 10 meters long is leaning against a vertical wall with its other end on the ground. The top end of the log is sliding down the wall. When the top end is 6 meters from the ground, it slides down at 2m/sec. How fast is the bottom moving away from the wall at this instant?
Answer:
Step-by-step explanation:
This is a related rates problem from calculus using implicit differentiation. The main equation is Pythagorean's Theorem. Basically, what we are looking for is [tex]\frac{dx}{dt}[/tex] when y = 6 and [tex]\frac{dy}{dt}=-2[/tex].
The equation for Pythagorean's Theorem is
[tex]x^2+y^2=c^2[/tex] where x and y are the legs and c is the hypotenuse. The length of the hypotenuse is 10, so when we find the derivative of this function with respect to time, and using implicit differentiation, we get:
[tex]2x\frac{dx}{dt}+2y\frac{dy}{dt}=0[/tex] and divide everything by 2 to simplify:
[tex]x\frac{dx}{dt}+y\frac{dy}{dt}=0[/tex]. Looking at that equation, it looks like we need a value for x, y, [tex]\frac{dx}{dt}[/tex] and [tex]\frac{dy}{dt}[/tex].
Since we are looking for [tex]\frac{dx}{dt}[/tex], that can be our only unknown and everything else has to have a value. So what do we know?
If we construct a right triangle with 10 as the hypotenuse and use 6 for y, we can solve for x (which is the only unknown we have, actually). Using Pythagorean's Theorem to solve for x:
[tex]x^2+6^2=10^2[/tex] and
[tex]x^2+36=100[/tex] and
[tex]x^2=64[/tex] so
x = 8.
NOW we can fill in the derivative and solve for [tex]\frac{dx}{dt}[/tex].
Remember the derivative is
[tex]x\frac{dx}{dt}+y\frac{dy}{dt}=0[/tex] so
[tex]8\frac{dx}{dt}+6(-2)=0[/tex] and
[tex]8\frac{dx}{dt}-12=0[/tex] and
[tex]8\frac{dx}{dt}=12[/tex] so
[tex]\frac{dx}{dt}=\frac{12}{8}=\frac{6}{4}=\frac{3}{2}=1.5 m/sec[/tex]
Find the Taylor series for f(x) centered at the given value of a. (Assume that f has a power series expansion. Do not show that Rn(x)→0 . f(x)=lnx, a=
Answer:
Here we just want to find the Taylor series for f(x) = ln(x), centered at the value of a (which we do not know).
Remember that the general Taylor expansion is:
[tex]f(x) = f(a) + f'(a)*(x - a) + \frac{1}{2!}*f''(a)(x -a)^2 + ...[/tex]
for our function we have:
f'(x) = 1/x
f''(x) = -1/x^2
f'''(x) = (1/2)*(1/x^3)
this is enough, now just let's write the series:
[tex]f(x) = ln(a) + \frac{1}{a} *(x - a) - \frac{1}{2!} *\frac{1}{a^2} *(x - a)^2 + \frac{1}{3!} *\frac{1}{2*a^3} *(x - a)^3 + ....[/tex]
This is the Taylor series to 3rd degree, you just need to change the value of a for the required value.
What is the value of x?
Which statement is true regarding the functions on the
graph?
f(6) = g(3)
f(3) = g(3)
f(3) = g(6)
f(6) = g(6)
Answer:
f(3) = g(3)
Step-by-step explanation:
on the graph the only point, where both lines cross (both functions create the same functional value) is at x=3.
since both lines have the same y-value there, we express this in math by the "=" sign. and both functions have the same input value (x=3) there.
Student received 10 different resistors for a laboratory setup with five slots to attach resistors, where each slot can accommodate only one resistor. In how many ways those 10 resistors can be attached to the laboratory setup?
Answer:
The number of ways of attaching the 10 resistors = 5¹⁰ = 9,765,625 ways
Step-by-step explanation:
Given;
total number of resistors, n = 10
number of slots available, = 5
The first resistor can be attached in 5 ways,
The second resistor can also be attached in 5 ways,
The third resistor can also be attached in 5 ways, etc
Each of the resistors can be attached in 5 different ways;
The number of ways of attaching the 10 resistors = 5¹⁰ = 9,765,625 ways
Find m < A
Round to the nearest degree.
CA = 6
CB = 13
AB = 10
Given the following numbers: a = 12500000 b = 0.00125 c = 1120000
Calculate (ab)÷ (c) and write the answer in standard form. (2.5 marks)
d) Express the interval (-1.5, 4] as an inequality and then graph the interval.
Answer:
Answer to the following question is as follows.
Step-by-step explanation:
Given:
a = 12500000
b = 0.00125
c = 1120000
Calculate (ab) ÷ (c)
Given:
d) Express the interval [-1.5, 4] as an inequality and then graph
Computation:
(ab) ÷ (c) = (a)(b) / c
(ab) ÷ (c) = (12500000)(0.00125) / (1120000)
(ab) ÷ (c) = 25 / 1,792
Express the interval [-1.5, 4]
{x : -1.5 < x ≤ 4}
Graph.
._________._________.
-1.5 0 4
Find the equation of the line passing through (3,5) with a slope of 1
WILL GIVE BRAINLIEST
Slope-intercept form:
y = x + 2
Point-slope form:
y − 5 = 1 ⋅ ( x − 3 )
I hope this is correct and helps!
Which line is parallel to the line that passes through the points (1,7) and (-3, 4)? A. y=--x-5 B. y=+*+1 y=-x-8 O c. D. 11 v==x+3 4
Answer:
B
Step-by-step explanation:
because
An initial deposit of $212 is placed in
a bank account and left to grow, with
interest compounded continuously.
what will it be after 6 years?
Round your answer to the nearest dollar.
Answer:
$224.932
Step-by-step explanation:
Note: The question is not complete
say the rate is 10%
Given data
Initial depostite= $212
TIme= 6years
rate= 10%
the expression for the compound interest is given as
A=P(1+r)^t
substitute
A=212(1+0.1)^6
A=212(1.01)^6
A=212*1.061
A= $224.932
Hence the final amount at the rate of 10% is $224.932
You own a small storefront retail business and are interested in determining the average amount of money a typical customer spends per visit to your store. You take a random sample over the course of a month for 12 customers and find that the average dollar amount spent per transaction per customer is $116.194 with a standard deviation of $11.3781. Create a 90% confidence interval for the true average spent for all customers per transaction.1) ( 114.398 , 117.99 )2) ( 112.909 , 119.479 )3) ( -110.295 , 122.093 )4) ( 110.341 , 122.047 )5) ( 110.295 , 122.093 )
Answer:
(110.295, 122.093).
Step-by-step explanation:
We have the standard deviation for the sample, which means that the t-distribution is used to solve this question.
The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So
df = 12 - 1 = 11
90% confidence interval
Now, we have to find a value of T, which is found looking at the t table, with 11 degrees of freedom(y-axis) and a confidence level of [tex]1 - \frac{1 - 0.9}{2} = 0.95[/tex]. So we have T = 1.7959
The margin of error is:
[tex]M = T\frac{s}{\sqrt{n}} = 1.7959\frac{11.3781}{\sqrt{12}} = 5.899[/tex]
In which s is the standard deviation of the sample and n is the size of the sample.
The lower end of the interval is the sample mean subtracted by M. So it is 116.194 - 5.899 = 110.295
The upper end of the interval is the sample mean added to M. So it is 116.194 + 5.899 = 122.093
So
(110.295, 122.093).
The radius of a sphere is increasing at a rate of 2 mm/s. How fast is the volume increasing (in mm3/s) when the diameter is 100 mm
Answer:
The radius is increasing at a rate of 62832 cubic millimeters per second when the diameter is of 100 mm.
Step-by-step explanation:
Volume of a sphere:
The volume of a sphere of radius r is given by:
[tex]V = \frac{4\pi r^3}{3}[/tex]
How fast is the volume increasing:
To find this, we have to differentiate the variables of the problem, which are V and r, implicitly in function of time. So
[tex]\frac{dV}{dt} = 4\pi r^2\frac{dr}{dt}[/tex]
The radius of a sphere is increasing at a rate of 2 mm/s.
This means that [tex]\frac{dr}{dt} = 2[/tex]
How fast is the volume increasing (in mm3/s) when the diameter is 100 mm?
Radius is half the diameter, so [tex]r = \frac{100}{2} = 50[/tex]
Then
[tex]\frac{dV}{dt} = 4\pi r^2\frac{dr}{dt} = 4\pi (50)^2(2) = 62832[/tex]
The radius is increasing at a rate of 62832 cubic millimeters per second when the diameter is of 100 mm.
question:
A sequence is defined by the recursive function f(n + 1) = –10f(n).
If f(1) = 1, what is f(3)?
3
–30
100
–1,000
the answer is 100
Answer:
100
Step-by-step explanation:
f(1) = 1
f(2) = -10×f(1) = -10 × 1 = -10
f(3) = -10×f(2) = -10 × -10 × f(1) = -10 × -10 × 1 = 100
f(n) = -10 to the power of n-1
Answer:
c - 100
Step-by-step explanation:
express the ratio as a fraction in the lowest term.3600s:2hours
Step-by-step explanation:
3600s=1hr
so, 1hr:1hr
1:1
Based on the concept of fractions and the information in the question, the fraction form in the lowest term is 1/2.
What is Fraction?Fraction is a term that is used to describe the portion/part of the whole thing. It represents the equal parts of the whole.
Generally, the term fraction has two parts, namely numerator and denominator.
Hence, in this case, to express the ratio as a fraction in its lowest term, convert both units to the same unit of time.
1 hour is equal to 3600 seconds so 2 hours is equal to 2 * 3600 = 7200 seconds.
Now the ratio is 3600 seconds to 7200 seconds.
To simplify this ratio we can divide both terms by their greatest common divisor which is 3600.
So the simplified ratio is 1:2.
Therefore, in this case, it is concluded that the fraction form in the lowest term is 1/2.
Learn more about fraction here: https://brainly.com/question/30154928
#SPJ2
This graph shows the solution to which inequality?
(32)
(-3.-6);
A ys 1/x - 2
B. y> fx-2
C. yzfx-2
***-2
Decide which of the two given prices is the better deal and explain why.
You can buy shampoo in a 5-ounce bottle for 3,89$ or in a 14-ounce bottle for 11,99$.
Select the correct choice below and, if necessary, fill in the answer box to complete your choice.
A.The -ounce bottle is the better deal because the cost per ounce is $
nothing per ounce while the -ounce bottle is $
nothing per ounce.
B.The -ounce bottle is the better deal because the cost per ounce is $
nothing per ounce while the -ounce bottle is $
nothing per ounce.
(Round to the nearest cent as needed.)
Answer:
The 14-ounce bottle is the better deal
Step-by-step explanation:
I know this beause inorder to figure out which one is better you have to make them the same price and then see which bottle has more ounces. So I made each price 1$ so there is 1.58-ounces per dollar in the 5-ounce bottle and 1.17 -ounces per dollar in the 14-ounce bottle.
what’s the answer to this problem please and thank you
Answer:
1.8574 hours
Step-by-step explanation:
Solve for t.
Take the natural log of both sides.
[tex] 3000 = 75000e^{-1.733t} [/tex]
[tex] 1 = 25e^{-1.733t} [/tex]
[tex] \dfrac{1}{25} = e^{-1.733t} [/tex]
[tex] \ln \dfrac{1}{25} = \ln (e^{-1.733t}) [/tex]
[tex] -3.218875 = -1.733t [/tex]
[tex] t = 1.8574[/tex]
Find the union {6, 11, 15} U Ø
Explanation:
The Ø means "empty set". It's the set with nothing inside it, not even 0.
We can write Ø as { } which is a pair of curly braces with nothing between them.
The rule is that if we union any set A with Ø, then we'll get set A
A U Ø = A
Ø U A = A
In a sense, it's analogous to adding 0. So it's like saying A+0 = A and 0+A = A.
So that's why {6, 11, 15} U Ø = {6, 11, 15}
There's nothing to add onto the set {6, 11, 15}, so we just get the same thing back again.
Which expression is equivalent to -9x-1y-9/-15x5y-3?
Answer: -9x-1y-9/
Step-by-step explanation:
Answer: b
Step-by-step explanation:
I really dont like edge
Find the radius of a circle with a diameter whose endpoints are (-7,1) and (1,3).
Answer:
r = 4.1231055
Step-by-step explanation:
So to do this, you need to find the distance between the two points:
(-7,1) and (1,3).
To do this, the distance or diameter (d) is equal to:
d = sqrt ((x2-x1)^2 + (y2-y1)^2)
In this case:
d = sqrt( (1 - (-7))^2 + (3 - 1)^2 )
d = sqrt( 8^2 + 2^2)
d = sqrt( 64 + 4)
d = sqrt( 68 )
The radius is half of the diameter, so:
r = 1/2 * d
r = 1/2 * sqrt( 68 )
r~ 4.1231055
HELP HELP HELPPPP
ILL GIVE BRAINLIEST HELPPPPPPPPP
100 POINTSSS
Answer:
C. 0.48
Step-by-step explanation:
Probability = number of required outcome
_______________________
number of possible outcome
= total volleyball game events
_______________________
total sophomore + junior
= 66/137
= 0.48
Answer: D) 0.31
Step-by-step explanation:
Let A denote the event that a person is a sophomore.
Let B denote the event that a person has attended volleyball game.
A∩B denote the event that a person is a sophomore and attend volleyball game.
Let P denote the probability of an event.
We are asked to find:
P(A∩B)
From the table provided to us we see that:
A∩B=42
Hence,
P(A∩B)=42/137=0.3065 which is approximately equal to 0.31. Therefore ur answer will be 0.31.
what is the greatest common factor of 160 and 198?
Hey there!
[tex]\large\textsf{FACTORS OF 160: 1, 2, 4, 5, 8, 10, 16, 20, 32, 40, 80, \& 160}[/tex]
[tex]\large\textsf{FACTORS OF 198: 1, 2, 3, 6, 9, 11, 18, 22, 33, 66, 99, \& 198}[/tex]
[tex]\large\text{Go through the factors to see if there’s any like terms and if you find any,} \\\large\text{look for the greatest one the numbers share together.}[/tex]
[tex]\large\text{Like terms: \boxed{\textsf{\bf 1 \& 2}}}[/tex]
[tex]\large\checkmark\boxed{\large\text{GCF: \bf 2 }}\large\checkmark[/tex]
[tex]\boxed{\boxed{\large\textsf{Answer: \huge the GCF \underline{G}reatest \underline{C}ommon \underline{F}actor is \bf 2}}}\huge\checkmark[/tex]
[tex]\large\textsf{Good luck on your assignment and enjoy your day!}[/tex]
~[tex]\frak{Amphitrite1040:)}[/tex]
If f(x)=5x and g(x)=2x-1, what is the composition f(g(x))?
Answer:
10x-5
Step-by-step explanation:
f(x)=5x
g(x)=2x-1
To create a composite function, replace x in f(x) with g(x)
f(g(x)) = 5(g(x) = 5(2x-1) = 10x-5
A professor is interested in whether or not college students have a preference (indicated by a satisfaction score) for reading a textbook that has a layout of one column or layout of two columns. In the above experiment, what is the dependent variable
Answer:
Satisfaction score
Step-by-step explanation:
The dependent variable may be described as the variable which is being measured in a research experiment. In the scenario described above, the dependent variable is the satisfaction score which is used to measure preference for a one or two column textbook. The dependent variable can also seen as the variable which we would like to predict, also called the predicted variable . The predicted variable here is the satisfaction score.
Find the point P along the directed line segment from point A(–9, 5) to point B(11, –2) that divides the segment in the ratio 4 to 1.
Answer:
[tex]P = (7, -\frac{3}{5})[/tex]
Step-by-step explanation:
Given
[tex]A = (-9,5)[/tex]
[tex]B = (11,-2)[/tex]
[tex]m : n = 4 : 1[/tex]
Required
Point P
This is calculated as:
[tex]P = (\frac{m * x_2 + n * x_1}{m + n}, \frac{m * y_2 + n * y_1}{m + n})[/tex]
So, we have:
[tex]P = (\frac{4 * 11 + 1 * -9}{4 + 1}, \frac{4 * -2 + 1 * 5}{4+1})[/tex]
[tex]P = (\frac{35}{5}, \frac{-3}{5})[/tex]
[tex]P = (7, -\frac{3}{5})[/tex]
Young invested GH150,000 and 2.5% per annum simple interest. how long will it take this amount to. yield an interest of GH11,250,00
Answer: 3 years
Step-by-step explanation:
Interest is calculated as:
= (P × R × T) / 100
where
P = principal = 150,000
R = rate = 2.5%.
I = interest = 11250
T = time = unknown.
I = (P × R × T) / 100
11250 = (150000 × 2.5 × T)/100
Cross multiply
1125000 = 375000T
T = 1125000/375000
T = 3
The time taken will be 3 years
Estimate 620 / 17 by first rounding each number so that it has only 1 nonzero digit.
no links plz
Step-by-step explanation:
620 / 17 =36.47058.. ≈ 36.5
Problem: The height, X, of all 3-year-old females is approximately normally distributed with mean 38.72
inches and standard deviation 3.17 inches. Compute the probability that a simple random sample of size n=
10 results in a sample mean greater than 40 inches. That is, compute P(mean >40).
Gestation period The length of human pregnancies is approximately normally distributed with mean u = 266
days and standard deviation o = 16 days.
Tagged
Math
1. What is the probability a randomly selected pregnancy lasts less than 260 days?
2. What is the probability that a random sample of 20 pregnancies has a mean gestation period of 260 days
or less?
3. What is the probability that a random sample of 50 pregnancies has a mean gestation period of 260 days
or less?
4. What is the probability a random sample of size 15 will have a mean gestation period within 10 days of
the mean?
Know
Learn
Booste
V See
Answer:
0.1003 = 10.03% probability that a simple random sample of size n= 10 results in a sample mean greater than 40 inches.
Gestation periods:
1) 0.3539 = 35.39% probability a randomly selected pregnancy lasts less than 260 days.
2) 0.0465 = 4.65% probability that a random sample of 20 pregnancies has a mean gestation period of 260 days or less.
3) 0.004 = 0.4% probability that a random sample of 50 pregnancies has a mean gestation period of 260 days or less.
4) 0.9844 = 98.44% probability a random sample of size 15 will have a mean gestation period within 10 days of the mean.
Step-by-step explanation:
To solve these questions, we need to understand the normal probability distribution and the central limit theorem.
Normal Probability Distribution
Problems of normal distributions can be solved using the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.
Central Limit Theorem
The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean
For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.
The height, X, of all 3-year-old females is approximately normally distributed with mean 38.72 inches and standard deviation 3.17 inches.
This means that [tex]\mu = 38.72, \sigma = 3.17[/tex]
Sample of 10:
This means that [tex]n = 10, s = \frac{3.17}{\sqrt{10}}[/tex]
Compute the probability that a simple random sample of size n= 10 results in a sample mean greater than 40 inches.
This is 1 subtracted by the p-value of Z when X = 40. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
By the Central Limit Theorem
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{40 - 38.72}{\frac{3.17}{\sqrt{10}}}[/tex]
[tex]Z = 1.28[/tex]
[tex]Z = 1.28[/tex] has a p-value of 0.8997
1 - 0.8997 = 0.1003
0.1003 = 10.03% probability that a simple random sample of size n= 10 results in a sample mean greater than 40 inches.
Gestation periods:
[tex]\mu = 266, \sigma = 16[/tex]
1. What is the probability a randomly selected pregnancy lasts less than 260 days?
This is the p-value of Z when X = 260. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{260 - 266}{16}[/tex]
[tex]Z = -0.375[/tex]
[tex]Z = -0.375[/tex] has a p-value of 0.3539.
0.3539 = 35.39% probability a randomly selected pregnancy lasts less than 260 days.
2. What is the probability that a random sample of 20 pregnancies has a mean gestation period of 260 days or less?
Now [tex]n = 20[/tex], so:
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{260 - 266}{\frac{16}{\sqrt{20}}}[/tex]
[tex]Z = -1.68[/tex]
[tex]Z = -1.68[/tex] has a p-value of 0.0465.
0.0465 = 4.65% probability that a random sample of 20 pregnancies has a mean gestation period of 260 days or less.
3. What is the probability that a random sample of 50 pregnancies has a mean gestation period of 260 days or less?
Now [tex]n = 50[/tex], so:
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{260 - 266}{\frac{16}{\sqrt{50}}}[/tex]
[tex]Z = -2.65[/tex]
[tex]Z = -2.65[/tex] has a p-value of 0.0040.
0.004 = 0.4% probability that a random sample of 50 pregnancies has a mean gestation period of 260 days or less.
4. What is the probability a random sample of size 15 will have a mean gestation period within 10 days of the mean?
Sample of size 15 means that [tex]n = 15[/tex]. This probability is the p-value of Z when X = 276 subtracted by the p-value of Z when X = 256.
X = 276
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{276 - 266}{\frac{16}{\sqrt{15}}}[/tex]
[tex]Z = 2.42[/tex]
[tex]Z = 2.42[/tex] has a p-value of 0.9922.
X = 256
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{256 - 266}{\frac{16}{\sqrt{15}}}[/tex]
[tex]Z = -2.42[/tex]
[tex]Z = -2.42[/tex] has a p-value of 0.0078.
0.9922 - 0.0078 = 0.9844
0.9844 = 98.44% probability a random sample of size 15 will have a mean gestation period within 10 days of the mean.
