Answer :
We are given that the weight [tex]$X$[/tex] of a 40‐year-old man is normally distributed with mean [tex]$\mu = 147$[/tex] pounds and standard deviation [tex]$\sigma = 16$[/tex] pounds. That is,
[tex]$$
X \sim N(147,16^2).
$$[/tex]
We wish to find the probability
[tex]$$
P(120 \le X \le 153).
$$[/tex]
A common strategy is to standardize the variable [tex]$X$[/tex] to a standard normal variable [tex]$Z$[/tex], defined by
[tex]$$
Z = \frac{X-\mu}{\sigma}.
$$[/tex]
### Step 1. Standardize the Endpoints
For the lower endpoint [tex]$X = 120$[/tex]:
[tex]$$
z_1 = \frac{120-147}{16} = \frac{-27}{16} \approx -1.6875.
$$[/tex]
For the upper endpoint [tex]$X = 153$[/tex]:
[tex]$$
z_2 = \frac{153-147}{16} = \frac{6}{16} = 0.375.
$$[/tex]
### Step 2. Express the Probability in Terms of [tex]$Z$[/tex]
The probability that [tex]$X$[/tex] is between 120 and 153 is the same as the probability that [tex]$Z$[/tex] is between [tex]$z_1$[/tex] and [tex]$z_2$[/tex]. Hence,
[tex]$$
P(120 \le X \le 153) = P(z_1 \le Z \le z_2) = \Phi(z_2) - \Phi(z_1),
$$[/tex]
where [tex]$\Phi(z)$[/tex] denotes the cumulative distribution function (CDF) of the standard normal distribution.
### Step 3. Use the Standard Normal Distribution Values
From the standard normal distribution table or an appropriate calculator, we find:
- [tex]$\Phi(z_1) \approx 0.0458$[/tex], when [tex]$z_1\approx -1.6875$[/tex].
- [tex]$\Phi(z_2) \approx 0.6462$[/tex], when [tex]$z_2=0.375$[/tex].
### Step 4. Compute the Probability Difference
Subtracting the two values gives
[tex]$$
P(120 \le X \le 153) \approx 0.6462 - 0.0458 = 0.6004.
$$[/tex]
### Final Answer
Thus, the probability that a 40-year-old man weighs between 120 and 153 pounds is approximately
[tex]$$
\boxed{0.6004}.
$$[/tex]
[tex]$$
X \sim N(147,16^2).
$$[/tex]
We wish to find the probability
[tex]$$
P(120 \le X \le 153).
$$[/tex]
A common strategy is to standardize the variable [tex]$X$[/tex] to a standard normal variable [tex]$Z$[/tex], defined by
[tex]$$
Z = \frac{X-\mu}{\sigma}.
$$[/tex]
### Step 1. Standardize the Endpoints
For the lower endpoint [tex]$X = 120$[/tex]:
[tex]$$
z_1 = \frac{120-147}{16} = \frac{-27}{16} \approx -1.6875.
$$[/tex]
For the upper endpoint [tex]$X = 153$[/tex]:
[tex]$$
z_2 = \frac{153-147}{16} = \frac{6}{16} = 0.375.
$$[/tex]
### Step 2. Express the Probability in Terms of [tex]$Z$[/tex]
The probability that [tex]$X$[/tex] is between 120 and 153 is the same as the probability that [tex]$Z$[/tex] is between [tex]$z_1$[/tex] and [tex]$z_2$[/tex]. Hence,
[tex]$$
P(120 \le X \le 153) = P(z_1 \le Z \le z_2) = \Phi(z_2) - \Phi(z_1),
$$[/tex]
where [tex]$\Phi(z)$[/tex] denotes the cumulative distribution function (CDF) of the standard normal distribution.
### Step 3. Use the Standard Normal Distribution Values
From the standard normal distribution table or an appropriate calculator, we find:
- [tex]$\Phi(z_1) \approx 0.0458$[/tex], when [tex]$z_1\approx -1.6875$[/tex].
- [tex]$\Phi(z_2) \approx 0.6462$[/tex], when [tex]$z_2=0.375$[/tex].
### Step 4. Compute the Probability Difference
Subtracting the two values gives
[tex]$$
P(120 \le X \le 153) \approx 0.6462 - 0.0458 = 0.6004.
$$[/tex]
### Final Answer
Thus, the probability that a 40-year-old man weighs between 120 and 153 pounds is approximately
[tex]$$
\boxed{0.6004}.
$$[/tex]
