In: Math
4. In a normal distribution with ?(?) = 100 and ??(?) = 16, find the predicted a. 8th percentile. b. 22nd percentile. c. median. d. ?3. e. 95th percentile.
6 In a data set, ? = 38, ?̅= 110.4, and ? = 20.9. In a normal distribution having the same features, find a. the predicted 30th percentile. b. the predicted 70th percentile. c. the predicted percentile corresponding to the 20th ordered data value. d. the predicted percentile corresponding to the 6th ordered data value.
4)
a)µ= 100
σ = 16
proportion= 0.0800
Z value at 0.08 =
-1.4051 (excel formula =NORMSINV(
0.08 ) )
z=(x-µ)/σ
so, X=zσ+µ= -1.405 *
16 + 100
X = 77.5189
(answer)
b)
µ= 100
σ = 16
proportion= 0.22
Z value at 0.22 =
-0.7722 (excel formula =NORMSINV(
0.22 ) )
z=(x-µ)/σ
so, X=zσ+µ= -0.772 *
16 + 100
X = 87.64 (answer)
c)
µ= 100
σ = 16
proportion= 0.5
Z value at 0.5 =
0.0000 (excel formula =NORMSINV(
0.5 ) )
z=(x-µ)/σ
so, X=zσ+µ= 0.000 *
16 + 100
X = 100.00 (answer)
d)
µ= 100
σ = 16
proportion= 0.75
Z value at 0.75 =
0.6745 (excel formula =NORMSINV(
0.75 ) )
z=(x-µ)/σ
so, X=zσ+µ= 0.674 *
16 + 100
X = 110.79 (answer)
e)
µ= 100
σ = 16
proportion= 0.95
Z value at 0.95 =
1.6449 (excel formula =NORMSINV(
0.95 ) )
z=(x-µ)/σ
so, X=zσ+µ= 1.645 *
16 + 100
X = 126.32 (answer)
============
6)
a)
µ = 110.4
σ = 20.9
n= 38
proportion= 0.3000
Z value at 0.3 =
-0.524 (excel formula =NORMSINV(
0.30 ) )
z=(x-µ)/(σ/√n)
so, X=z * σ/√n +µ= -0.524 *
20.9 / √ 38 +
110.4 = 108.622
b)
µ = 110.4
σ = 20.9
n= 38
proportion= 0.7000
Z value at 0.7 =
0.524 (excel formula =NORMSINV(
0.70 ) )
z=(x-µ)/(σ/√n)
so, X=z * σ/√n +µ= 0.524 *
20.9 / √ 38 +
110.4 = 112.178
c)
µ = 110.4
σ = 20.9
n= 38
proportion= 0.2000
Z value at 0.2 =
-0.842 (excel formula =NORMSINV(
0.20 ) )
z=(x-µ)/(σ/√n)
so, X=z * σ/√n +µ= -0.842 *
20.9 / √ 38 +
110.4 = 107.547
d)
µ = 110.4
σ = 20.9
n= 38
proportion= 0.0600
Z value at 0.06 =
-1.555 (excel formula =NORMSINV(
0.06 ) )
z=(x-µ)/(σ/√n)
so, X=z * σ/√n +µ= -1.555 *
20.9 / √ 38 +
110.4 = 105.129