Asample of size 40 is taken from an infinite population whose mean and standard deviation are 68 and 12 respectively. the probability that the sample mean is larger than 70 equals:

To solve this problem, we can use the z statistic to find for the probability. The formula for z score is:

z = (x – u) / s

Where,

u = the sample mean = 68

s = standard deviation of the samples = 12

x = sample value = 70

In this case, what is asked is the probability that the sample mean is larger than 70, therefore this corresponds to the right of z on the graph of normal distribution.

z = (70 – 68) / 12

z = 2 / 12

z = 0.17

Therefore finding for P at z ≥ 0.17 using the standard distribution tables:

P (z = 0.17) = 0.5675

But this is not the answer yet since this is the P to the left of z. Therefore the correct one is:

P (z ≥ 0.17) = 1 - 0.5675

P (z ≥ 0.17) = 0.4325 = 43.25%

The probability that the sample mean is larger than 70 is 43.25%

step-by-step explanation:   divide the numerator by the denominator to get the decimal

it would still be 193

if divide by 100 then 1.93

i am not sure

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