Statistics quesion re: sample distribution?

Question

I need help, with explanation, please for the following question:

The tensile strength of spot welds produced by a robot welder is normally distributed, with a mena of 10,000 pounds per square inch and a standard deviation of 800 pounds per square inch. For a simple random sample of n = 4 welds, what is the probability that the sample mean will be at least 10,200 pounds per square inch? Less than 9900 pounds per square inch?

Answer 1

ANSWER: PROBABILITY ABOVE 10,200 LBS/IN^2  is (approx) 40%

ANSWER: PROBABILITY BELOW 9,900 LBS/IN^2 is (approx) 45%

Why??

--------ABOVE 10,200 LBS/IN^2

NORMAL DISTRIBUTION, STANDARDIZED VARIABLE z, PROBABILITY "LOOK-UP"

z = (x - μ)/σ [0.25]

z = STANDARDIZED VARIABLE FOR NORMAL DISTRIBUTION

x = INDEPENDENT VARIABLE [10200]

μ = POPULATION MEAN [10000]

σ = POPULATION STANDARD DEVIATION [800]

NORMAL DISTRIBUTION (CUMULATIVE) "LOOK-UP" Table for z = 0.25 is (approx) 0.60   (60%)

PROBABILITY = 1 - 0.60 = 0.40 (40%)

--------BELOW 9,900 LBS/IN^2

NORMAL DISTRIBUTION, STANDARDIZED VARIABLE z, PROBABILITY "LOOK-UP"

z = (x - μ)/σ [-0.125]

NORMAL DISTRIBUTION (CUMULATIVE) "LOOK-UP" Table for z = -0.125 is (approx) 0.45 (45%)