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

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

answer: idk

step-by-step explanation:

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step-by-step explanation: