Question

The braking distance for a Krazy-Car traveling at 50 mph is normally distributed with a mean of 50 ft. and a standard deviation of 5 ft. Answer the following without using the z table.

a. What is the random variable in this problem? _______________________________

b. If top 3.5% of the cars are going too fast, what would be the braking distance?

c. Find Q1, the median, Q3 and IQR. ________________________________

Answer 1

(a)

the random variable in this problem : The braking distance for a Krazy-Car traveling at 50 mph

(b)

= 50

= 5

Top 3.5% corresponds to Cumulative Area Under Standard Normal Curve = 1 - 0.035 = 0.965

By Technology, corresponding Z score = 1.81191

So we get:

Z = 1.81191 = ( X - 50)/5

So,

X = 50 + ( 1.81191 X 5)

= 59.05955

So,

Answer is:

59.05955

(c)

(i)

Q1 corresponds to Cumulative Area Under Standard Normal Curve = 0.25

By Technology, corresponding Z score = - 0.67448975

So we get:

Z = - 0.67448975 = ( X - 50)/5

So,

X = 50 - ( 0.67448975 X 5)

= 46.62755125

So,

Answer is:

46.62755125

(ii)

Median = Mean = 50

So,

Answer is:

50

(iii)

Q3 corresponds to Cumulative Area Under Standard Normal Curve = 0.75

By Technology, corresponding Z score = 0.67448975

So we get:

Z = 0.67448975 = ( X - 50)/5

So,

X = 50 + ( 0.67448975 X 5)

= 53.37244875

So,

Answer is:

53.37244875

(iv)

IQR = Q3 - Q1 = 53.37244875 - 46.62755125 = 6.7448975

So,

Answer is:

6.7448975