Answer to Question #115640 in Statistics and Probability for Wale

Question

An agency is conducting performance tests on different brands of skateboards.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 will require at least a pressure of 70 to break?
c) In what range centered on the mean will 80% of the data lie?

Answer 1

Let x be the random variable for pressure.

X~N(54.7, 6.2^{2 })

a) P(50<X<60)

$Z=\frac {X-\mu} {\sigma}$

$Z_1=\frac{50-54.7}{6.2}$

=-0.7581

$Z_2=\frac{60-54.7}{6.2}$

=0.8548

P(50<X<60)=$\Phi (Z_2)-\Phi (Z_1)$

=$\Phi (0.8548)-\Phi (-0.7581)$

Read the corresponding values from a standard normal table

=0.80234-0.22363

=0.57871

b)$P(X\geq 70)$

=1-P(X<70)

Z=$\frac {70-54.7}{6.2}$ =2.4677

$\Phi (2.4677)=0.99324$

=1-0.99324

=0.00676

c) Range in which 80% of the data will lie

$\alpha =1-0.8 =0.2$

Since it is two sided, we divide 0.2 by 2 =0.1

$\Phi ^-(0.1) =1.2816$

-1.2816<z<1.2816

$±1.2816=\frac {X-54.7}{6.2}$

X=46.754, 62.646

80%of the data will lie between 46.754 and 62.646