In 2005, 0.76% of all airline flights were on-time. if we choose a simple random sample of 2000 flights, find the probability (to four decimal places, using normal chart, no continuity correction)

(a) at least 79% of the sample's flights were on time

(b) at most 1580 of the sample's flights were on time

(c) the sample proportion of on-time flights (p-hat) differs from the truth by more than three percent

Step-by-step explanation:

Here X no of flight that reach on time has two outcomes only and each trial is independent of the other.

Since sample size is large we approximate to normal with mean =np and variance =npq

i.e. X is N(152,19.10)

is used for converting to std normal variate

Continuity correction of 0.5 on either side of interval would be applied for x

a)

=0.00

b)=1.00

c) p is N(0.76, sqrtpq/n)=N(0.76,0.0095)

P(p diff>0.03) =

=0.00

option 2 is correct.

step-by-step explanation:

we are given two inequality equation

x+y> 5 and -3x+2y< -2

first we draw the graph of both function using t-table method

equation: x+y> 5

for x=0, y=5

for x=-5, y=0

table:

x   :   0     5

y   :   5     0

test point (0,0) , 0> 5

false

equation: -3x+2y< -2

for x=0, y=-1

for x=2, y=4

table:

x   :   0     2

y   : -1       4

test point (0,0) , 0< -2

true

now we draw the graph. see the attachment for graph.

common region would be solution set.

option 2 is correct.

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step-by-step explanation: