Look at the picture.
I have reached 250 thanks which means I helped 250 wonderful people :D
2 answers:
You might be interested in
Look at the picture.
First find the secant line. The slope of the secant line through 
(when 
) and 
(when 
) is the average rate of change of 
over the interval ![[-1,2]](https://tex.z-dn.net/?f=%5B-1%2C2%5D)
:
The tangent line to will have a slope determined by the derivative:
Both the secant and tangent will have the same slope when
, or when 
.
The tangent line to
Both the secant and tangent will have the same slope when
Answer:
a)
<em>The probability that a single score drawn at random will be greater than 110 </em>
<em>P( X > 110) = 0.1587</em>
<em>b) </em>
<em>The probability that a sample of 25 scores will have a mean greater than 105</em>
<em> P( x> 105) = 0.0062</em>
<em>c) </em>
<em>The probability that a sample of 64 scores will have a mean greater than 105</em>
<em> P( x⁻> 105) = 0.002</em>
<em></em>
<em>d) </em>
<em> The probability that the mean of a sample of 16 scores will be either less than 95 or greater than 105</em>
<em> P( 95 ≤ X≤ 105) = 0.9544</em>
<em></em>
Step-by-step explanation:
<u><em>a)</em></u>
Given mean of the Normal distribution 'μ' = 100
Given standard deviation of the Normal distribution 'σ' = 10
a)
Let 'X' be the random variable of the Normal distribution
let 'X' = 110
<em>The probability that a single score drawn at random will be greater than 110</em>
<em>P( X > 110) = P( Z >1)</em>
= 1 - P( Z < 1)
= 1 - ( 0.5 +A(1))
= 0.5 - A(1)
= 0.5 -0.3413
= 0.1587
b)
let 'X' = 105
<em>The probability that a single score drawn at random will be greater than 110</em>
<em> P( x> 105) = P( z > 2.5)</em>
<em> = 1 - P( Z< 2.5)</em>
<em> = 1 - ( 0.5 + A( 2.5))</em>
<em> = 0.5 - A ( 2.5)</em>
<em> = 0.5 - 0.4938</em>
<em> = 0.0062</em>
<em>The probability that a single score drawn at random will be greater than 105</em>
<em> P( x> 105) = 0.0062</em>
<em>c) </em>
let 'X' = 105
<em>The probability that a single score drawn at random will have a mean greater than 105</em>
<em> P( x> 105) = P( z > 4)</em>
<em> = 1 - P( Z< 4)</em>
<em> = 1 - ( 0.5 + A( 4))</em>
<em> = 0.5 - A ( 4)</em>
<em> = 0.5 - 0.498</em>
<em> = 0.002</em>
<em> The probability that a sample of 64 scores will have a mean greater than 105</em>
<em> P( x⁻> 105) = 0.002</em>
<em>d) </em>
<em>Let x₁ = 95</em>
<em>Let x₂ = 105</em>
The probability that the mean of a sample of 16 scores will be either less than 95 or greater than 105
P( 95 ≤ X≤ 105) = P( -2≤z≤2)
= P(z≤2) - P(z≤-2)
= 0.5 + A( 2) - ( 0.5 - A( -2))
= A( 2) + A(-2) (∵A(-2) =A(2)
= A( 2) + A(2)
= 2 × A(2)
= 2×0.4772
= 0.9544
<em> The probability that the mean of a sample of 16 scores will be either less than 95 or greater than 105</em>
<em> P( 95 ≤ X≤ 105) = 0.9544</em>
<em> </em>