Answer to Question #118635 in Statistics and Probability for Michael

Question

An agency is conducting performance tests on different brands of skateboard. Tests show that the pressure, in pounds per square inch, required to break the boards is normally distributed with a mean of 54.7 and a standard deviation of 6.2

a. What percent of the boards have breaking pressures between 50 and 60?
b. What is the probability that any board will require at least a pressure of 70 to break ?
c. In what range centred on the mean will 80% of the data lie?

Answer 1

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($\frac{50-54.7}{6.2}\leq\frac{X-54.7}{6.2}\leq\frac{60-54.7}{6.2}$)

= 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($\frac{X-54.7}{6.2}\geq\frac{70-54.7}{6.2}$)

= 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. $2\Phi(m/6.2)$ = 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).