Activity 2
1. In a Science test, the mean score is 42 and the standard deviation is 5.
Assuming the scores are normally distributed, what percent of the score is
a. Greater than 48?
b. Less than 50?
c. Between 30 and 48?
Sketch the normal
curve
μ=42,σ=5,Z=(x−μ)σ\mu=42,\sigma=5,Z=\frac{(x-μ)}{σ}μ=42,σ=5,Z=σ(x−μ)
a. P(x>48)=P(Z>48−425)=P(Z>1.2)P(x>48)=P(Z>\frac{48-42}{5})=P(Z>1.2)P(x>48)=P(Z>548−42)=P(Z>1.2)
=1−P(Z<1.2)=1-P(Z<1.2)=1−P(Z<1.2)
=1−0.8849=0.1151=1-0.8849=0.1151=1−0.8849=0.1151
=11.51=11.51=11.51 %
b. P(x<50)=P(Z<50−425)=P(Z<1.6)P(x<50)=P(Z<\frac{50-42}{5})=P(Z<1.6)P(x<50)=P(Z<550−42)=P(Z<1.6)
P(Z<1.6)=0.9452P(Z<1.6)=0.9452P(Z<1.6)=0.9452
=94.52=94.52=94.52 %
c. P(30<x<48)=P(30−425<Z<48−425)=P(−2.4<Z<1.2)P(30<x<48)=P(\frac{30-42}{5}<Z<\frac{48-42}{5})=P(-2.4<Z<1.2)P(30<x<48)=P(530−42<Z<548−42)=P(−2.4<Z<1.2)
=P(Z<1.2)−P(Z<−2.4)=P(Z<1.2)-P(Z<-2.4)=P(Z<1.2)−P(Z<−2.4)
=0.8849−0.0082=0.8849-0.0082=0.8849−0.0082
=0.8767=0.8767=0.8767
=87.67=87.67=87.67 %
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