Consider the probability that no less than 99 out of 150 cell phone calls will not be disconnected. Assume the probability that a given cell phone call will not be disconnected is 63%. Approximate the probability using the normal distribution. Round your answer to four decimal places.

The probability that no less than 99 out of 150 cell phone calls will not be disconnected = P(x > 99) = 0.2240

Step-by-step explanation:

The mean = μ = np = 150 × 0.63 = 94.5

The standard deviation of the sample is given as √(np(1-p))

σ = √(150×0.63×0.37) = 5.91

For normal distribution, we will need to normalize the given variable whose probability we seek.

The standardized score for any value is the value minus the mean then divided by the standard deviation.

z = (x - μ)/σ = (99 - 94.5)/5.91 = 0.76

To determine the probability that no less than 99 out of 150 cell phone calls will not be disconnected = P(x > 99) = P(z > 0.76)

We'll use data from the normal probability table for these probabilities

P(x > 99) = P(z > 0.76) = 1 - P(z ≤ 0.76) = 1 - 0.776 = 0.2240

option (b) is correct

step-by-step explanation:

to find the solution of this system, we need to convert it into reduced row-echelon form. we have,

after applying row operations r2 -> r2/2 and r3 -> r3/3, we get,