meredith48034
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the weights of cars passing over a bridge has a mean of 3550 lb and a standard deviation of 870 lb. Assume that the weights of the cars are passing over the bridge are normally distributed. What is the probability that thr weight of a randomly-selected car passing over the bridge is than 3,000 pounds.

a. 50.15%
b. 25.93%
c. 47.13%
d. 74.07%​

Respuesta :

mathmate

Step-by-step explanation:

Note: Question does not indicate if probability required is for weight to exceed or below 3000 lbs.  So choose appropriate answer accordingly (near the end)

Using the usual notations and formulas,

mean, mu = 3550

standard deviation, sigma = 870

Observed value, X = 3000

We calculate

Z = (X-mu)/sigma = (3000-3550)/870 = -0.6321839

Probability of weight below 3000 lbs

= P(X<3000) = P(z<Z) = P(z<-0.6321839) = 0.2636334

Answer:

Probability that a car randomly selected is less than 3000

= P(X<3000) = 0.2636 (to 4 decimals)

Probability that a car randomly selected is greater than 3000

= 1 - P(X<3000) = 1 - 0.2636 (to 4 decimals) = 0.7364 (to 4 decimals)

Answer:

d is the closest if you meant more than

b is the closet if you meant less than

Step-by-step explanation:

We need to figure out the z-number.

The z-number is computed by:

[tex]\frac{x-\mu}{\sigma}[/tex] where [tex]\mu[/tex] is mean and [tex]\sigma[/tex] is standard deviation.

[tex]\frac{3000-3550}{870}=\frac{-550}{870}=-0.63218[/tex] approximately.

So P(X<3000)=P(Z<-0.63218)

Since this is normally distribute P(Z<-0.63218) is the same as P(Z>0.63218).

To find P(Z>0.63218) you must compute 1-P(Z<0.63218).

P(X<3000)=P(Z<-0.63218)

                =1-P(Z<0.63218)

To find this we need to find the row for 0.6 and the column for .03 since 0.6+.03 is 0.63

               =1-.7357

               =.2643

As a percentage this is 26.43%.

The closet choice is b. 25.93%

P(x>3000)=1-P(x<3000)=1-.2643=.7357 or 73.57%.

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