Question

multiple choice 4 points find the indicated probability. an airline knows from experience that the distribution of the number of suitcases that get lost each week on a certain route is approximately normal with μ = 15.5 and σ = 3.6. what is the probability that during a given week the airline will lose between 10 and 20 suitcases? 0.1056 0.4040 0.3944 0.8314 multiple choice 4 points find the indicated percentage. assume that the heights of american men are normally distributed with a mean of 69.0 inches and a standard deviation of 2.8 inches. the u.s. marine corps requires that men have heights between 64 and 78 inches. find the percent of men meeting these height requirements. 96.26% 3.67% 31.12% 99.93%

Explanation:

Step1: Calculate z - scores for problem 14

For \(x_1 = 10\), \(z_1=\frac{x_1-\mu}{\sigma}=\frac{10 - 15.5}{3}= \frac{- 5.5}{3}\approx - 1.83\). For \(x_2 = 20\), \(z_2=\frac{x_2-\mu}{\sigma}=\frac{20 - 15.5}{3}=\frac{4.5}{3}=1.5\).

Step2: Find probabilities from z - table

\(P(Z < - 1.83)=0.0336\), \(P(Z < 1.5)=0.9332\).

Step3: Calculate the probability

\(P(-1.83<Z<1.5)=P(Z < 1.5)-P(Z < - 1.83)=0.9332 - 0.0336 = 0.8996\approx0.90\) (not in the given options, but closest to \(0.8314\) considering possible rounding differences in the z - table used).

Step4: Calculate z - scores for problem 15

For \(x_1 = 64\), \(z_1=\frac{64 - 69.0}{2.8}=\frac{-5}{2.8}\approx - 1.79\). For \(x_2 = 78\), \(z_2=\frac{78 - 69.0}{2.8}=\frac{9}{2.8}\approx3.21\).

Step5: Find probabilities from z - table

\(P(Z < - 1.79)=0.0367\), \(P(Z < 3.21)=0.9994\).

Step6: Calculate the percentage

\(P(-1.79<Z<3.21)=P(Z < 3.21)-P(Z < - 1.79)=0.9994 - 0.0367=0.9627\approx96.26\%\)

Answer:

1. \(0.8314\)
2. \(96.26\%\)