Question

5)- The distribution of the demand (in number of units per unit time) for a product can often be approximated by a normal probability distribution. For example, a communication cable company has determined that the number of push-button terminal switches demanded daily has a normal distribution with mean 200 and standard deviation 50.

a) On what percentage of days will the demand be less than 90 switches?
b) On what percentage of days will the demand fall between 225 and 275 switches?
c) Based on cost considerations, the company has determined that its best strategy is to produce a sufficient number of switches so that it will fully supply demand on 94% of all days. How many terminal switches should the company produce per day?

Answer 1

a) On approximately 1.39% of days, the demand for switches will be less than 90. b) The probability of demand falling between 225 and 275 switches is 24.17%. c) The company should produce approximately 278 terminal switches per day to fully supply demand on 94% of all days.

Step-by-step explanation:

a) To calculate the percentage of days when the demand will be less than 90 switches, we need to find the probability of demand being less than 90 switches using the normal distribution. We can use the z-score formula to calculate the z-score for 90 switches:

z = (x - mean) / standard deviation

z = (90 - 200) / 50 = -2.2

Using a standard normal distribution table or calculator, we can find that the probability of z < -2.2 is approximately 0.0139 or 1.39%.

So, on approximately 1.39% of days, the demand for switches will be less than 90.

b) To calculate the percentage of days when the demand will fall between 225 and 275 switches, we need to find the probabilities of demand being less than 275 switches and demand being less than 225 switches, and then subtract the two probabilities. Using the z-score formula:

z1 = (225 - 200) / 50 = 0.5

z2 = (275 - 200) / 50 = 1.5

Using a standard normal distribution table or calculator, we can find that the probability of z < 0.5 is approximately 0.6915 and the probability of z < 1.5 is approximately 0.9332. So, the probability of demand falling between 225 and 275 switches is 0.9332 - 0.6915 = 0.2417 or 24.17%.

c) To find the number of terminal switches the company should produce per day to fully supply demand on 94% of all days:

We need to find the z-score that corresponds to a cumulative probability of 0.94. Using a standard normal distribution table or calculator, we can find that the z-score is approximately 1.5548.

Now, we can use the z-score formula to find the number of switches:

z = (x - mean) / standard deviation

1.5548 = (x - 200) / 50

Solving for x, we get:

x = 1.5548 * 50 + 200 = 277.74

So, the company should produce approximately 278 terminal switches per day to fully supply demand on 94% of all days.