91Ó°ÊÓ

The lengths of foots for women is normally distributed with mean\(\left( \mu \right)\)9.6 in. and standard deviation is 0.5 in\(\left( \sigma \right)\).

a) Let X be the foot length of women.

Then,

\[\begin{aligned}{c}X \sim N\left( {\mu ,{\sigma ^2}} \right)\\ \sim N\left( {9.6,0.{5^2}} \right)\end{aligned}\]

The value for z-score is,

\(\begin{aligned}{c}z = \frac{{x - \mu }}{\sigma }\\ = \frac{{10 - 9.6}}{{0.5}}\\ = 0.8\end{aligned}\)

The probability that a randomly selected woman has a foot length less than 10.0 in.is given by\(P\left( {X < 10.00} \right) = P\left( {Z < 0.8} \right)\).

Refer to the standard normal table for the cumulative value of 0.800 as 0.7881.Thus, \[P\left( {Z < 0.8} \right) = 0.7881\].

Therefore, the probability that a randomly selected woman has a foot length less than 10.0 in. is 0.7881.

b) The z-scores corresponding to the foots lengths are:

\(\begin{aligned}{c}{z_1} = \frac{{{x_1} - \mu }}{\sigma }\\ = \frac{{8.0 - 9.6}}{{0.5}}\\ = - 3.2\end{aligned}\)

\(\begin{aligned}{c}{z_2} = \frac{{{x_2} - \mu }}{\sigma }\\ = \frac{{11.0 - 9.6}}{{0.5}}\\ = 2.8\end{aligned}\)

The probability that a randomly selected woman has a foot length between 8.0 in. and 11.0 in. is given by,

\(\begin{aligned}{c}P\left( {8.0 < X < 11.0} \right) = P\left( { - 3.2 < Z < 2.8} \right)\\ = P\left( {Z < 2.8} \right) - P\left( {Z < - 3.2} \right)\;\;\;\;\;\;\;\;\;...\left( 1 \right)\end{aligned}\)

Refer to the standardnormal table.

The cumulative area corresponding to 2.8is 0.9974.

The cumulative area corresponding to -3.2, is 0.0007.

Substitute the values in equation (1),

\(\begin{aligned}{c}P\left( { - 3.2 < z < 2.8} \right) = P\left( {z < 2.8} \right) - P\left( {z < - 3.2} \right)\\ = 0.9974 - 0.0007\\ = 0.9967\end{aligned}\)

Therefore, the probability that a randomly selected woman with foot lengths between 8.0 in. and 11.0 in. is 0.9967.

c) Let the\({P_{95}}\)be the 95^th percentile of foot lengths and z be the corresponding z-score.

\(\begin{aligned}{c}P\left( {X < {P_{95}}} \right) = 0.95\\P\left( {Z < z} \right) = 0.95\end{aligned}\)

Where,\(z = \frac{{{P_{95}} - \mu }}{\sigma }\).

The z-score for\({P_{95}}\)from the table is 1.645.

Percentile is given by,

\(\begin{aligned}{c}z = 1.645\\{P_{95}} = \mu + \left( {1.645 \times \sigma } \right)\\ = 9.6 + \left( {1.645 \times 0.5} \right)\\ = 10.4\;{\rm{in}}\end{aligned}\)

Thus, the 95^th percentile is 10.4 in.

d) Given that sample size is 25(n),

Let \(\bar X\) be the sample mean distribution for foot lengths for 25 women.

As the population is normally distributed, the sample mean distribution would be normal.

\(\begin{aligned}{c}{\mu _{\bar X}} = \mu \\ = 9.6\;{\rm{in}}\end{aligned}\)

And

\(\begin{aligned}{c}{\sigma _{\bar X}} = \frac{\sigma }{{\sqrt n }}\\ = \frac{{0.5}}{{\sqrt {25} }}\\ = 0.1\;{\rm{in}}\end{aligned}\)

Thus,

\(\begin{aligned}{c}\bar X \sim N\left( {{\mu _{\bar X}},{\sigma _{\bar X}}^2} \right)\\ \sim N\left( {9.6,{{0.1}^2}} \right)\end{aligned}\)

The sample mean value of 9.8 in. has corresponding z-score as,

\(\begin{aligned}{c}z = \frac{{\bar x - {\mu _{\bar X}}}}{{{\sigma _{\bar X}}}}\\ = \frac{{9.8 - 9.6}}{{0.1}}\\ = 2\end{aligned}\)

The probability that 25 women have foot lengths with a mean greater than 9.8 in. is given by,

\(\begin{aligned}{c}P\left( {\bar X > 9.8} \right) = P\left( {Z > 2} \right)\\ = 1 - P\left( {Z < 2} \right)\\ = 1 - 0.9772\\ = 0.0228\end{aligned}\)

Refer to standard normal distribution to obtain the left tailed area corresponding to the value 2.00, which is 0.0228.