Question #198576

5) It is known that amounts of money spent on clothing in a year by students on a particular campus follow a normal distribution with a mean of $380 and a standard deviation of $50. a. What is the probability that a randomly chosen student will spend less than $400 on clothing in a year? b. What is the probability that a randomly chosen student will spend more than $360 on clothing in a year? c. What is the probability that a randomly chosen student will spend between $300 and $400 on clothing in a year?


Expert's answer

Solution:

μ=380,σ=50X∼N(μ,σ)\mu=380,\sigma=50 \\X\sim N(\mu,\sigma)

(a) P(X<400)=P(z<400−38050)=P(z<0.4)=0.65542P(X<400)=P(z<\dfrac{400-380}{50})=P(z<0.4)=0.65542

(b) P(X>360)=1−P(X≤360)=1−P(z≤360−38050)P(X>360)=1-P(X\le360)=1-P(z\le\dfrac{360-380}{50})

=1−P(z≤−0.4)=1−[P(z≥0.4)]=1−[1−P(z≤0.4)]=P(z≤0.4)=0.65542=1-P(z\le-0.4)=1-[P(z\ge0.4)]=1-[1-P(z\le0.4)] \\=P(z\le0.4)=0.65542

(c) P(300<X<400)=P(X<400)−P(X<300)P(300<X<400)=P(X<400)-P(X<300)

=P(z<400−38050)−P(z<300−38050)=P(z<0.4)−P(z<−1.6)=P(z<0.4)−[1−P(z≤1.6)]=0.65542−1+0.94520=0.60062=P(z<\dfrac{400-380}{50})-P(z<\dfrac{300-380}{50}) \\=P(z<0.4)-P(z<-1.6) \\=P(z<0.4)-[1-P(z\le1.6)] \\=0.65542-1+0.94520 \\=0.60062


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