Suppose that a random sample of size 36 is to be selected from a population with mean 50 and standard deviation 7. What is the approximate probability that X will be within .5 of the population mean

Step-by-step explanation:

Let us assume that x is normally distributed. The sample size is greater than 30. Since the the population mean and population standard deviation are known, we would apply the formula,

z = (x - µ)/(σ/√n)

Where

x = sample mean

µ = population mean

σ = standard deviation

n = number of samples

From the information given,

µ = 50

σ = 7

n = 36

If x is within 0.5 of the population mean, it means that x is between (50 - 0.5) and (50 + 0.5)

the probability is expressed as

P(49.5 ≤ x ≤ 50.5)

For x = 49.5

z = (49.5 - 50)/(7/√36) = - 0.43

Looking at the normal distribution table, the probability corresponding to the z score is 0.334

For x = 49.5

z = (50.5 - 50)/(7/√36) = 0.43

Looking at the normal distribution table, the probability corresponding to the z score is 0.666

Therefore,

P(49.5 ≤ x ≤ 50.5) = 0.666 - 0.334 = 0.332

554.4

step-by-step explanation:

44*12.6=544.4