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Z-SCORES3. Suppose the speeds of cars on a street are normally distributed,with a mean of 76.4 kph and a standard deviation of 5.0 kph.Part A: To the nearest tenth of a percent, what is the probabilitythat a car is going slower than 70 kph? (3 points: 2 points forcorrectly finding the probability and 1 point for expressing theanswer to a tenth of a percent)Part B: To the nearest tenth of a percent, what is the probabilitythat a car is going faster than 80 kph? (4 points: 3 points forcorrectly finding the probability and 1 point for expressing theanswer to a tenth of a percent)Part C: To the nearest tenth of a percent, what is the probabilitythat a car is going between 72 kph and 78 kph? (7 points: 6points for correctly finding the probability and 1 point forexpressing the answer to a tenth of a percent)

ZSCORES3 Suppose The Speeds Of Cars On A Street Are Normally Distributedwith A Mean Of 764 Kph And A Standard Deviation Of 50 KphPart A To The Nearest Tenth Of class=

Sagot :

The Solution.

Z-score formula is given as below:

[tex]Z=\frac{x-\mu}{\sigma}[/tex]

In this case,

[tex]x=70,\mu=76.4,\sigma=5[/tex]

Therefore,

[tex]Z=\frac{70-76.4}{5}=-\frac{6.4}{5}=-1.28[/tex]

Hence, using the negative z-score table, we have

[tex]Pr(Z<-1.28)=0.1003[/tex]

Hence, the probability for part A is 10.0%

For part B:

First, we find the z-score of 80

[tex]\begin{gathered} \text{ the value of x now becomes} \\ x=80 \\ \text{ thus} \\ z=\frac{80-76.4}{5}=\frac{3.6}{5}=0.72 \end{gathered}[/tex]

Hence, using the positive z-score table, we have

[tex]Pr(Z>0.72)=1-Pr(Z\le0.72)=1-0.7642=0.2358[/tex]

Therefore, the probability for part B is 23.6%

For part C:

First, we find the z-score of 72 and 78

[tex]\begin{gathered} when\text{ the value of x becomes} \\ x=72 \\ \text{ thus} \\ z=\frac{72-76.4}{5}=\frac{-4.4}{5}=-0.88 \end{gathered}[/tex]

[tex]\begin{gathered} when\text{ the value of x becomes} \\ x=78 \\ \text{ thus} \\ z=\frac{78-76.4}{5}=\frac{1.6}{5}=0.32 \end{gathered}[/tex]

Hence, using the positive z-score table, we have

[tex]Pr(-0.88

Therefore, the probability for part C is 43.6%

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