Question

suppose that the weight of seedless watermelons is normally distributed with mean 6.2 kg. and standard deviation 1.5 kg.
a. what is the z - score for a seedless watermelon weighing 8 kg?
select

b. what is the probability that a randomly selected watermelon will weigh more than 7 kg? select

c. what is the probability that a randomly selected seedless watermelon will weigh between 4 and 5 kg? select

Explanation:

Step1: Calculate z-score (Part A)

The z-score formula is $z = \frac{x - \mu}{\sigma}$, where $x=8$, $\mu=6.2$, $\sigma=1.5$.
$z = \frac{8 - 6.2}{1.5} = \frac{1.8}{1.5} = 1.2$

Step2: Find $P(X>7)$ (Part B)

First calculate z-score for $x=7$:
$z = \frac{7 - 6.2}{1.5} \approx 0.53$
Find $P(Z>0.53) = 1 - P(Z \leq 0.53)$. From z-table, $P(Z \leq 0.53) \approx 0.7019$, so $1 - 0.7019 = 0.2981$

Step3: Find $P(4

Calculate z-scores for $x=4$ and $x=5$:
$z_1 = \frac{4 - 6.2}{1.5} \approx -1.47$, $z_2 = \frac{5 - 6.2}{1.5} = -0.8$
Find $P(-1.47 < Z < -0.8) = P(Z < -0.8) - P(Z < -1.47)$. From z-table, $P(Z < -0.8)=0.2119$, $P(Z < -1.47)\approx0.0708$, so $0.2119 - 0.0708 = 0.1411$

Answer:

A. 1.2
B. ~0.298 (or 0.2981)
C. ~0.141 (or 0.1411)