# Answer in Statistics and Probability for Michael #118635

Let X = the random variable denoting the pressure required to break the skateboards

By the problem we have,

X ~ N(“mu” = 54.7, “sigma”2 = 6.22)

Then,

Z = “frac{X-mu}{sigma}” ~ N(0, 1), Z is called the standard normal variable

a. To find the percent of the boards have breaking pressures between 50 and 60 we find

the probability of the boards have breaking pressures between 50 and 60

= P(50 “leq” X “leq” 60)

= P(“frac{50-54.7}{6.2}leqfrac{X-54.7}{6.2}leqfrac{60-54.7}{6.2}”)

= P(- 0.76 “leq” Z “leq” 0.85)

= “Phi”(0.85) – “Phi”(-0.76)

= 0.8023 – 0.2236

= 0.5787

Answer: The percent of the boards have breaking pressures between 50 and 60 = 0.5787 x 100 = 57.9.

b. The probability that any board will require at least a pressure of 70 to break

= P(X “geq” 70)

= P(“frac{X-54.7}{6.2}geqfrac{70-54.7}{6.2}”)

= P(Z “geq” 2.47)

= 1 – P(Z < 2.47)

= 1 – “Phi”(2.47)

= 1 – 0.9932

= 0.0068

Answer: The probability that any board will require at least a pressure of 70 to break is 0.0068.

c. Let m be the value on the either side of the mean within which 80% of the data lie

Then, P(“mu” – m “leq” X “leq” “mu” + m) = 80%

i.e. P(- m/6.2 “leq” Z “leq” m/6.2) = 80%

i.e. “Phi(m/6.2)-Phi(-m/6.2)” = 80%

i.e. 2“Phi(m/6.2)-1” = 80%

i.e. “2Phi(m/6.2)” = 1.8

i.e. “Phi(m/6.2)” = 0.9 = “Phi(1.28)”

i.e. m = 1.28 x 6.2 = 7.94

Answer: Within +/- 7.94 pounds per square inch range centred on the mean will 80% of the data lie i.e. within (mean – 7.94 to mean + 7.94).

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