The average time to run the 5K fun run is 22 minutes and the standard deviation is 2.2 minutes. 44 runners are randomly selected to run the 5K fun run. Round all answers to 4 decimal places where possible and assume a normal distribution.
a)X ~ N(22,2.2)
b)
sample size =n= | 44 |
std error=σx̅=σ/√n= | 0.3317 |
x¯ ~ N(22 , 0.3317)
c)
total expected value=22*44 =968
standard deviation =2.2*sqrt(44)=14.5931
∑x ~ N(968 ,14.5931)
d)
probability =P(21.9024<X<22.2024)=P((21.9024-22)/2.2)<Z<(22.2024-22)/2.2)=P(-0.04<Z<0.09)=0.5367-0.4823=0.0543 |
e)
probability =P(21.9024<X<22.2024)=P((21.9024-22)/0.332)<Z<(22.2024-22)/0.332)=P(-0.29<Z<0.61)=0.7292-0.3843=0.3449 |
f)
probability =P(X<985.6)=(Z<985.6-968)/14.593)=P(Z<(1.206)=0.8861 |
g)
No
h)
for 20th percentile critical value of z= | -0.842 | ||
therefore corresponding value=mean+z*std deviation= | 955.7181 |
Get Answers For Free
Most questions answered within 1 hours.