in a survey of 18 - year - old males, the mean weight was 154.9 pounds with a standard deviation of 45.4…

Explanation:

Step1: Recall the z - score formula

The formula for a value (x) in a normal distribution is (x=\mu + z\sigma), where (\mu) is the mean, (z) is the z - score, and (\sigma) is the standard deviation. Given (\mu = 154.9) pounds and (\sigma=45.4) pounds.

Step2: Find the z - score for the 94th percentile

Using a standard normal distribution table (or z - table), the z - score (z) corresponding to a cumulative probability of (0.94) is approximately (z_{0.94}=1.55).

Step3: Calculate the weight for the 94th percentile

Substitute (\mu = 154.9), (z = 1.55), and (\sigma = 45.4) into the formula (x=\mu+z\sigma). So (x=154.9+1.55\times45.4). [ \begin{align*} x&=154.9 + 1.55\times45.4\ &=154.9+70.37\ &=225.27\approx225.3 \end{align*} ]

Step4: Find the z - score for the 42nd percentile

The cumulative probability is (0.42). Looking up in the z - table, the z - score (z_{0.42}\approx - 0.20).

Step5: Calculate the weight for the 42nd percentile

Substitute (\mu = 154.9), (z=- 0.20), and (\sigma = 45.4) into the formula (x=\mu + z\sigma). So (x=154.9+( - 0.20)\times45.4). [ \begin{align*} x&=154.9-0.20\times45.4\ &=154.9 - 9.08\ &=145.82\approx145.8 \end{align*} ]

Step6: Recall the definition of the third - quartile

The third quartile (Q_3) represents the 75th percentile. The z - score (z) corresponding to a cumulative probability of (0.75) is approximately (z_{0.75}=0.67).

Step7: Calculate the weight for the third quartile

Substitute (\mu = 154.9), (z = 0.67), and (\sigma = 45.4) into the formula (x=\mu+z\sigma). So (x=154.9+0.67\times45.4). [ \begin{align*} x&=154.9+0.67\times45.4\ &=154.9 + 30.418\ &=185.318\approx185.3 \end{align*} ]

Answer:

(a) 225.3 pounds (b) 145.8 pounds (c) 185.3 pounds