Stat 133 Recitation Winter 2007
Additional Problems SOLUTIONS
Suppose the skid distance during an accident at this intersection has a normal
distribution with mean 50 feet and standard deviation 5 feet.
1.
What is the probability that one
randomly selected car skids more than 51 feet in an
accident at this intersection?
X has a normal distribution with mean = 50, std = 5
P(X>51) = P (Z > (5150)/5) = P (Z > 0.2) = 1 P(Z<0.2) = 10.5793 = 0.4207
2.
What is the probability that one randomly selected car skids between 46 and 51 feet in
an accident at this intersection?
Plot will help a lot!
P(46<X<51) = P(X<51) – P(X<46) = P(Z<(5150)/5) – P((4650)/5)
= P(Z<0.2) – P(Z<0.8) = 0.5793  0.2119 = 0.3674
3.
What is the 10
th
percentile for the skid distance for an accident at this intersection?
Want to find P(Z<?) = 0.1 ,
from Table A, Z = 1.28
Unstandardize
X= 50 + (1.28)*5 = 43.6
4.
What is the probability that the average
skid distance for a random sample of 100 cars
is more than 51 feet?
First find the distribution of sample mean Xbar
Xbar has a normal distribution with mean = 50, std = 5/sqrt(100)=0.5
P(Xbar>51) = P(Z > (5150)/0.5) = P(Z>2) = 1P(Z<2) = 1 0.9772 = 0.0228
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 Spring '07
 rumseyjohnson
 Statistics, Central Limit Theorem, Normal Distribution, Probability, Standard Deviation

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