19. the number of nails of a given length is normally distributed with a mean length of 5.00 in. and a…

Explanation:

Step1: Calculate z - scores

The formula for the z - score is $z=\frac{x-\mu}{\sigma}$, where $\mu = 5.00$ (mean), $\sigma=0.03$ (standard deviation), and $x$ is the value from the data set.

Step2: Solve part a

For $x = 4.94$, $z=\frac{4.94 - 5.00}{0.03}=\frac{- 0.06}{0.03}=-2$. Using the standard normal distribution table, the proportion of values less than $z = - 2$ is $0.0228$. The number of nails less than $4.94$ in a bag of $n = 120$ is $0.0228\times120 = 2.736\approx3$.

Step3: Solve part b

For $x = 4.97$, $z_1=\frac{4.97 - 5.00}{0.03}=\frac{-0.03}{0.03}=-1$. For $x = 5.03$, $z_2=\frac{5.03 - 5.00}{0.03}=\frac{0.03}{0.03}=1$. The proportion of values between $z=-1$ and $z = 1$ is $P(-1<Z<1)=\Phi(1)-\Phi(-1)$, where $\Phi(z)$ is the cumulative - distribution function of the standard normal distribution. From the standard normal table, $\Phi(1)=0.8413$ and $\Phi(-1)=0.1587$, so $P(-1 < Z < 1)=0.8413-0.1587 = 0.6826$. The number of nails between $4.97$ and $5.03$ in a bag of $120$ is $0.6826\times120 = 81.912\approx82$.

Step4: Solve part c

For $x = 5.03$, $z=\frac{5.03 - 5.00}{0.03}=1$. The proportion of values greater than $z = 1$ is $1-\Phi(1)=1 - 0.8413=0.1587$. The number of nails over $5.03$ in a bag of $120$ is $0.1587\times120 = 19.044\approx19$.

Answer:

a. 3 b. 82 c. 19