The number of pieces of mail that a department receives each day can be modeled by a distribution having mean 44 and standard deviation 8. for a random sample of 35 days, what can be said about the probability that the sample mean will be less than 40 or greater than 48 using?

Step-by-step explanation:

Assuming a normal distribution for the number of pieces of mail that a department receives each day, the formula for normal distribution is expressed as

z = (x - u)/s

Where

x = number of pieces of mail

u = mean

s = standard deviation

From the information given,

u = 44

s = 8

We want to find

P(x lesser than or 40) or P(x greater than 48)

For P(x lesser than or 40)

For x = 40,

z = (40 - 44)/8 = - 4/8 = - 0.5

Looking at the normal distribution table, the corresponding value of the z score is 0.30854

P(x greater than 48) = 1 - P(lesser than or or equal 48)

For x = 48,

z = (48 - 44)/8 = 4/8 = 0.5

Looking at the normal distribution table, the corresponding value of the z score is 0.69146

P(x greater than 48) = 1 - 0.69146 = 0.30854

The probability that the sample mean will be less than 40 or greater than 48 using is

0.30854 + 0.30854 = 0.69146

idk

step-by-step explanation: