Question #138213

1. Find the mean and the standard deviation of the heights of male engineering students in which the heights 66 inches and 75 inches correspond to standard scores of –0.6 and 1.1, respectively.

2. A traffic study of 1,600 vehicles that passed by a checkpoint showed that their speeds were normally distributed with a mean of 97.6 kph and a standard deviation of 10.4 kph. How many vehicles had a speed of more than 100 kph?

Expert's answer

1)Given  that,x1=66,x2=75,Z1=0.6,Z2=1.1,thenZ=xμσ0.6=66μσ    0.6σ=66μ,1.1=75μσ    1.1σ=75μ,by multiplying the first equation by (-1) and then adding the result to the second equation1.7σ=66+75=9    σ=91.75.294,μ=66+0.6(5.294)=69.17642)Given  that,μ=97.6,σ=10.4,then,a)P(x>100)=P(Z>10097..610.4)=P(Z>0.23)=0.5P(0<Z<0.23)=0.50.0910=0.409number of bulbs =1600×0.409=654.4655  vehicles1) Given \;that,\\ x_{1}=66,x_{2}=75,Z_{1}=-0.6,Z_{2}=1.1,then\\ Z=\frac{x-\mu}{\sigma}\\ \therefore -0.6=\frac{66-\mu}{\sigma}\implies -0.6\sigma=66-\mu,\\ \therefore 1.1=\frac{75-\mu}{\sigma}\implies 1.1\sigma=75-\mu,\\ \text{by multiplying the first equation by (-1) }\\ \text{and then adding the result to the second equation}\\ 1.7 \sigma=-66+75=9\implies \sigma=\frac{9}{1.7}\approx 5.294,\\ \mu=66+0.6(5.294)=69.1764\\ 2)Given \; that, μ=97.6, σ=10.4, then,\\ a) P(x>100) = P( Z >\frac{100-97..6}{10.4})\\ =P(Z>0.23)\\=0.5- P(0<Z<0.23)=0.5-0.0910=0.409\\ \text{number of bulbs }=1600\times 0.409\\ =654.4\approx655\;vehicles \\


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