An athletes score is distributed Uniformly on 0 10 for a sin

An athlete’s score is distributed Uniformly on (0, 10) for a single attempt. a). What is the probability that the score is between 4 and 6? b). Use CLT to compute the probability that the average score on 12 attempts is between 4 and 6? c). How many attempts are necessary so that the average score is between 4 and 6 with probability .99?

Solution

A)

Note that here,          
          
a = lower fence of the distribution =    0      
b = upper fence of the distribution =    10      
          
Thus, the area between the said numbers is          
          
c = lower number =    4      
d = higher number =    6      
          
Thus, the probability between these two values is          
          
P = (d - c)/(b - a) =    0.2   [ANSWER]

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b)

Thus, the mean, variance, and standard deviations are          
          
u = mean = (b + a)/2 =    5      
s^2 = variance = (b -a)^2 / 12 =    8.333333333      
s = standard deviation = sqrt(s^2) =    2.886751346

We first get the z score for the two values. As z = (x - u) sqrt(n) / s, then as          
x1 = lower bound =    4      
x2 = upper bound =    6      
u = mean =    5      
n = sample size =    12      
s = standard deviation =    2.886751346      
          
Thus, the two z scores are          
          
z1 = lower z score = (x1 - u) * sqrt(n) / s =    -1.2      
z2 = upper z score = (x2 - u) * sqrt(n) / s =    1.2      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.11506967      
P(z < z2) =    0.88493033      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.76986066   [ANSWER]

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c)

The z score necessary for the middle 99% is

z = 2.575829304

Thus,

(x1 - u) * sqrt(n) / s = 2.575829304

(6 - 5)*sqrt(n)/2.886751346 = 2.575829304

Solving for n,

n = 55.29080501 or approx 55 [ANSWER]