2. The normal curve is sketched below. Solve for z in the following problems. =7

Question

2. The normal curve is sketched below. Solve for z in the following problems. =77% 27, : 6o2. 3. A certain STAT 208 exam had scores that were normally distributed, with an average of 60 points and a standard deviation (SD) of 15 points. Find the percent of students who have scores (a) between 45 and 75 points (f) under 55 points b) between 30 and 90 points (g) over 50 poiits (c) between 50 and 65 points (h) under 65 points (d) between 65 and 95 points (i) ind the 25th percentile (e) over 90 pointe (J) find the 90th percentile

Explanation / Answer

a) P(-Z< x < Z) = 77% = 0.77

Two sided normal distribution table.

Total covered area =100%, 100-77 = 23 and 23/2 =11.5

to calculate Z value of P(x<Z) = 0.77+(1-0.77)/2 = 0.77+ 11.5 = 88.5

P(x<Z) =88.5, then Z= 1.2

P(-1.2 < Z <1.2) = 0.77 or 77%

Z= 1.2

b) P(-Z>x >Z) = 2*2 = 4%

96% confidence interval value;

0.96+0.02= 0.98

P(X<Z) = 0.98 and Z= 2.054

P(-2.054> X > 2.054) =4%

Z= 2.054

c) P(Z<x) = 0.60 or 60%

P(Z<0.254) = 0.60

Z= 0.254

d) P(x>Z) = 0.1 or 10%

P(X> Z) = 1- P(X<Z); P(x <Z) = 0.9

P(X<Z)= 0.9 then Z= 1.282

P(Z>1.282) =0.1

Z= 1.282

2) a) Mean= 45 and std= 15

Z= (x-Mean)/std

P(45 <x < 75) =P((45-60)/15 < Z < (75-60)/15) =P( -1 <Z <1) = 0.8413- 0.1587 = 0.6827

b) P(30 < x< 90) = P((30-60)/15 < Z <(90-60)/15) =P( -2 < Z< 2) =0.9772-0.0228 = 0.9545

c) P(50<X <65) = P((50-60) /15 <Z< (65-60)/15) =P(-0.6667<Z < 0.333) = 0.6304 - 0.2525 = 0.3779

d) P(65 <X< 95) =P((65-60)/15 < Z < (95-60)/15) = P(0.333 <Z< 2.333) = 0.9901- 0.6304 = 0.3597

e) P(X>90) = 1- P(X<90) = 1- P(Z< (90-60)/15)= 1- P(Z<2) = 1- 0.9772 = 0.0228

f) P(X<55) = P(Z< (55-60)/15)= P(Z< -0.3333)= 0.3695

g) P(X>50) = 1- P(X<50) = 1 - P(Z< (50-60)/15) = 1-0.2525= 0.7475

h) P( X<65) = P(Z<(65-60)/15) = P(Z<0.3333)= 0.6304

i) P(X=x) = 0.25

P(Z=X) =0.25

Z(0.25)= -0.674

Z= (X-mean)/ std => -0.674= (X-60)/15 => -10.11= X-60 => X= 60-10.11= 49.89

X=49.89

j) 90th percentile

P(X=x) = 0.9

P(Z=x) = 0.9

Z= 1.282

1.282= (X-60)/15 => X-60= 19.23 => X= 79.23

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