An engineer is going to redesign an ejection seat for an airplane. The seat was designed for pilots weighing between 140 lb and 191 lb. The new population of pilots has normally distributed weights with a mean of 148 lb and a standard deviation of 30.3 lb. a. If a pilot is randomly​ selected, find the probability that his weight is between 140 lb and 191 lb.

The probability is approximately . ​(Round to four decimal places as​ needed.)

The probability is approximately 0.3193

Step-by-step explanation:

The question parameters;

The mass between of the pilot for which the seat was designed, = 140 lb and 191 lb

The mean weight of the new pilots, μ = 148 lb

The standard deviation of the weights new pilots, s = 30.3 lb

a. The z-score is given as follows;

Therefore, we have;

The p-value = 0.60257

The p-value = 0.9222

The probability that the mean lie between the two values = 0.9222 - 0.60257 = 0.31963

Therefore, the probability that the weight is between the 140 lb and 191 lb ≈ 0.3196

- "runner b is moving at a slower rate."

- "the graphs shown are graphs of distance(km) against time(min).

the slopes of this graphs are are given by:

slope = (change in y)/(change in x)

= (change in distance)/(change in time)

this is the formula for finding the speed. so, the slope of these graphs represent the speed of the runner.

taking each division in the graph to represent 1 on both axis:

runner a ⇒ speed = (2-0)/(1-0)

= 2/1

= 2 km/min

runner b ⇒ speed = (3-0)/(3-0)

= 3/3

= 1 km/min

this shows that   runner b is moving at a slower rate."