Answer to Question #91571 in Statistics and Probability for fahad

Question

Q.The burning time of an experimental rocket is a random variable having the normal distribution with p = 4.76 second and a = 0.04 second. What is the probability that this kind
of rocket will bum.

i. Less than 4.66 seconds
II. More than 4.80 seconds
III. More Than From 4.70 to 4.82 seconds.

Answer 1

Answer

I. 0.0062,

II. 0.1562,

III. 0.8664.

Explanation

Let us use the common notations. The burning time of an experimental rocket obeys a normal probability law with mean µ=4.76 seconds and standard deviation s = 0.04 seconds.

We have a random variable X normally distributed with mean µ and standard deviation s. To solve the problems, we note that Z = (X – 4.76)/0.04 is the standardized version of X. So, we have

P(X < 4.66) = F_Z((4.66-4.76)/0.04),

P(X > 4.80) = 1 - P(X <= 4.80) = 1 - Fz((4.80-4.76)/0.04),

P(4.70 < X < 4.82) = F_Z((4.82-4.76)/0.04) - F_Z((4.70-4.76)/0.04),

where F_Z(x) is the cumulative distribution function for a standard normal random variable.

Calculating the arguments and using tabulated values of the cumulative distribution function for a standard normal distribution, we obtain

P(X < 4.66) = F_Z(–2.5) = 1 – F_Z(2.5) = 1 – 0.9938 = 0.0062,

P(X > 4.80) = 1 – F_Z(1.0) = 1 – 0.8413 = 0.1587,

P(4.70 < X < 4.82) = F_Z(1.5) – F_Z(–1.5) = F_Z(1.5) – (1 – F_Z(1.5)) = 0.8664.