Answer:
The probability of randomly selecting a rod that is shorter than 22 cm
P(X<22) = 0.1251
Step-by-step explanation:
<u><em>Step(i):</em></u>-
Given mean of the Population = 25cm
Given standard deviation of the Population = 2.60
Let 'x' be the random variable in normal distribution
Given x=22

<u><em>Step(ii):</em></u>-
The probability of randomly selecting a rod that is shorter than 22 cm
P(X<22) = P( Z<-1.15)
= 1-P(Z>1.15)
= 1-( 0.5+A(1.15)
= 0.5 - A(1.15)
= 0.5 - 0.3749
= 0.1251
The probability of randomly selecting a rod that is shorter than 22 cm
P(X<22) = 0.1251
Answer:
A) p =
d
Step-by-step explanation:
20(which is d) x
= 2 (which is p)
cross checking always helps!
Answer:
The probability is 
Step-by-step explanation:
From the question we are told that
The capacity of an Airliner is k = 300 passengers
The sample size n = 320 passengers
The probability the a randomly selected passenger shows up on to the airport

Generally the mean is mathematically represented as
=>
=>
Generally the standard deviation is

=> 
=> 
Applying Normal approximation of binomial distribution
Generally the probability that there will not be enough seats to accommodate all passengers is mathematically represented as

Here 
=>
Now applying continuity correction we have
=> ![P(X >300 ) = P(Z > \frac{[300.5] - 307.2}{3.50} )](https://tex.z-dn.net/?f=P%28X%20%20%3E300%20%29%20%3D%20%20P%28Z%20%3E%20%20%5Cfrac%7B%5B300.5%5D%20-%20307.2%7D%7B3.50%7D%20%29)
=> 
From the z-table

So

The answer is 1.72 and it would be 2 if u estimated