Question

the weight of oranges growing in an orchard is normally distributed with a mean weight of 8 oz. and a standard deviation of 1.5 oz. what percentage of the oranges from the orchard weigh less than 7 oz., to the nearest tenth? statistics calculator

Explanation:

Step1: Calculate the z-score

The formula for the z - score is $z=\frac{x - \mu}{\sigma}$, where $x$ is the value from the dataset, $\mu$ is the mean, and $\sigma$ is the standard deviation.
Given that $x = 7$, $\mu=8$, and $\sigma = 1.5$.
Substitute these values into the formula: $z=\frac{7 - 8}{1.5}=\frac{- 1}{1.5}\approx - 0.67$

Step2: Find the percentage using z - table or calculator

We want to find $P(X<7)$, which is equivalent to $P(Z < - 0.67)$ (where $Z$ is the standard normal variable).
Using a standard normal distribution table or a calculator, we know that the cumulative distribution function for the standard normal distribution gives us the probability that $Z$ is less than a given value.
Looking up $z=-0.67$ in the standard normal table or using a calculator (for example, in a TI - 84 plus, we can use the normalcdf function with lower bound $=-\infty$, upper bound $=- 0.67$, mean $=0$, standard deviation $=1$), we get $P(Z < - 0.67)\approx0.2514$

Step3: Convert to percentage and round

To convert the probability to a percentage, we multiply by 100: $0.2514\times100 = 25.14\%$
Rounding to the nearest tenth, we get $25.1\%$

Answer:

$25.1\%$