Answer:
Using z score formula:
X = z ∂ + µ
= 157.833
Step-by-step explanation:
Solution:
Mean = µ = 1262
Standard deviation = ∂ = 117
(a) 28th percentile for the number of chocolate chip.
P( z < z) = 28%
= 0.28
P( z<- 0.58) = 0.28
Z = -0.58
By using z score formula:
Z = x - µ /∂
-0.58= x – 117 / 1262
X = (- 0.58)(117) + (1262)
= 1194.14
(b) Middle 97% of bag.
P(-z < z < z) = 97%
= 0.97
P( z < z) – p(z < -z) = 0.97
2p(z < z) -1 = 0.97
2p (z < z) = 1 + 0.97
P(z < z) = 1.97 / 2
= 0.99
P(z < 2.33) = 0.99
Z ± 2.33
By using z score formula:
Z = x - µ / ∂
X = z ∂ + µ
= - 2.33 x 117 + 1262
=989.39
Z = 2.33
X = z ∂ + µ
= 2.33 x 117 + 1262
=1533.61
(c) What is the interquartile range of the number of chocolate chips in a bag of chocolate.
By using standard normal table,
The z dist’n formula:
P(z < z ) = 25%
=0.25
P(z < -0.6745) = 0.25
Z = 0.6745
Using z score formula:
X = z∂ + µ
= - 0.6745 x 117 + 1262
= 1183.0835
First quartile = Q1 =1183.0835
The third quartile is:
P(z<z) = 75%
= 0.75
P(z < 0.6745) = 0.75
Z = 0.6745
Using z score formula:
X = z ∂ + µ
= 0.6745 x 117 + 1262
= 1340.9165
IQR = Q3 – Q1
= 1340.9165 – 1183.0835
= 157.833
Answer:
i cant see te pic that good
Step-by-step explanation:
Answer:
5/12
Step-by-step explanation:
Just took quiz and got problem right.
Answer: x = -3 y = 2 So (-3,2)
Step-by-step explanation:
- x + 3y = 9 Multiply everything by 2 then add to the second equation.
2x + 5y = 4
<u>-2x +6y =18 </u> then 2x-2x=0 Add the other terms and solve for y.
11y = 22 divide both sides by 11 yo get y = 2 Substitute that into either one and solve for x.
2x + 5y = 4
2x + 5(2) = 4
2x + 10 = 4 subtract 10 from both sides 2x = 4 -10
2x = -6 so divide both sides by 2 to get x = -3
