Answer:
(a) 
(b) 
(c) P(X ≤ 1) = 0.6321
(d) P(2 ≤ X ≤ 5) = 0.1286
Step-by-step explanation:
From the given information; Let consider X to be the time between two successive arrivals at the drive-up window of a local bank.
However; If X is regarded as the exponential distribution with λ = 1 which is identical to a standard gamma distribution with ∝ = 1
The objective is to compute the following :
(a) The expected time between two successive arrivals is;



(b) The standard deviation of the time between successive arrivals is;




(c) P(X ≤ 1)
P(X ≤ 1) = 1 - e⁻¹
P(X ≤ 1) = 0.6321
(d) P(2 ≤ X ≤ 5)
P(2 ≤ X ≤ 5) = [1 - e⁻⁵] - [1 - e⁻²]
P(2 ≤ X ≤ 5) = 0.9933 - 0.8647
P(2 ≤ X ≤ 5) = 0.1286
-2x
6 - 4 = 2
BOTH NUMBERS HAVE VARIABLE OF X SO THE X WILL GO ON THE 2 ALSO
6x - 4x = 2x
The value of the product expression is –6p^3 + 8p^2 – 10p
<h3>How to simplify the product?</h3>
The product expression is given as:
2p(–3p2 + 4p – 5)
Rewrite properly as:
2p(–3p^2 + 4p – 5)
Remove the bracket
2p * –3p^2 + 2p * 4p – 2p * 5
Evaluate the products
–6p^3 + 8p^2 – 10p
Hence, the value of the product expression is –6p^3 + 8p^2 – 10p
Read more about expression at
brainly.com/question/4541471
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