Ask Question
6 September, 14:32

Before a new phone system was installed, the amount a company spent on personal calls followed a normal distribution with an average of $900 per month and a standard deviation of $50 per month. Refer to such expenses as PCE's (personal call expenses). Using the distribution above, what is the probability that during a randomly selected month PCE's were between $775.00 and $990.00

0
Answers (1)
  1. 6 September, 15:07
    0
    Answer: the probability that during a randomly selected month, PCE's were between $775.00 and $990.00 is 0.9538

    Step-by-step explanation:

    Since the amount that the company spent on personal calls followed a normal distribution, then according to the central limit theorem,

    z = (x - µ) / σ

    Where

    x = sample mean

    µ = population mean

    σ = standard deviation

    From the information given,

    µ = $900

    σ = $50

    the probability that during a randomly selected month PCE's were between $775.00 and $990.00 is expressed as

    P (775 ≤ x ≤ 990)

    For (775 ≤ x),

    z = (775 - 900) / 50 = - 2.5

    Looking at the normal distribution table, the probability corresponding to the z score is 0.0062

    For (x ≤ 990),

    z = (990 - 900) / 50 = 1.8

    Looking at the normal distribution table, the probability corresponding to the z score is 0.96

    Therefore,

    P (775 ≤ x ≤ 990) = 0.96 - 0.0062 = 0.9538
Know the Answer?
Not Sure About the Answer?
Get an answer to your question ✅ “Before a new phone system was installed, the amount a company spent on personal calls followed a normal distribution with an average of ...” in 📙 Mathematics if there is no answer or all answers are wrong, use a search bar and try to find the answer among similar questions.
Search for Other Answers