Answer to Question #198576 in Statistics and Probability for fahmida

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?


1
Expert's answer
2021-05-31T19:17:29-0400

Solution:

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

(a) P(X<400)=P(z<40038050)=P(z<0.4)=0.65542P(X<400)=P(z<\dfrac{400-380}{50})=P(z<0.4)=0.65542

(b) P(X>360)=1P(X360)=1P(z36038050)P(X>360)=1-P(X\le360)=1-P(z\le\dfrac{360-380}{50})

=1P(z0.4)=1[P(z0.4)]=1[1P(z0.4)]=P(z0.4)=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<40038050)P(z<30038050)=P(z<0.4)P(z<1.6)=P(z<0.4)[1P(z1.6)]=0.655421+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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