Question

3 pts Question 9 Astudy involving stress is conducted among the students on a college campus. The stress scores follow auniform distribution with the lowest stress score equal to one and the highest equal to five. Usinga sample of 55 students, find: a. Find Pl< 7).Draw the graph b.Find P(Ex 170). c. Find the 80th percentile for the mean of 55 scores. d. Find the 85th percentile for the sum of 55 scores Since the individual stress scores follow a uniform distribution, X-U(1, 5) where a -1 and b 5 (See Continuous Random Variables e for an explanation on the uniform distribution). Upload Choose a File

Answer 1

Let x be the stress score of the students. x follows uniform distribution with a = 1 and b = 5

Therefore area between 1 to 5 would be 1 and area outside 1 to 5 would be 0

with Mean ( µ ) =  
a b 2
= 3 and Variance ( σ² ) =
(b a)2 12
= 1.3333

1/4 7 0 1 5 7

a) P ( x < 7 )  

According to probability density function .the area between 0 to 1 is 0 ; the area between 1 to 5 is 1 and area between 5 to 7 is 0

Therefore P ( x < 7 ) = 0+1+0 = 1

b) P( ∑x > 170 )

  ∑x follows approximately normal distribution with mean n*µ = 55*3 = 165 and standard deviation = = 8.5634

2 Vn*o

=

170 165 2г - теап P sd 8.5634

= P( z > 0.58 )

= 1 - P ( z ≤ 0.58 )

= 1 - 0.7190 ------ ( from z score table )

= 0.2810

c) Mean of 55 scores ( $\bar{x}$ ) follows approximately normal distribution with $\mu _\bar{x}$ = µ = 3 and

$\sigma _\bar{x}$ = $\sqrt{\frac{\sigma ^2}{n}}$ = 0.1557

We are asked to find 80th percentile for $\bar{x}$ , that is we have to find $\bar{x}$ such that area to the left is 0.80

Therefore we need to find z corresponding to area 0.80 on z score table , so such z score = 0.84

0.80 z-0.84

$\bar{x}$ = z*$\sigma _\bar{x}$ + $\mu _\bar{x}$ = ( 0.84*0.1557) + 3  

$\bar{x}$ = 3.13

d)

∑x follows approximately normal distribution with mean n*µ = 55*3 = 165 and standard deviation = = 8.5634

2 Vn*o

We are asked to find 85th percentile for ∑x , that is we have to find ∑x such that area to the left is 0.85

Therefore we need to find z corresponding to area 0.85 on z score table , so such z score = 1.04

0.85 z 1.04

∑x = z*Mean + SD = (1.04*165 ) + 8.5634

∑x = 173.91