question\nthe weight of oranges growing in an orchard is normally distributed with a mean weight of 7.5 oz…
question\nthe weight of oranges growing in an orchard is normally distributed with a mean weight of 7.5 oz. and a standard deviation of 1.5 oz. from a batch of 2000 oranges, how many would be expected to weight more than 5 oz., to the nearest whole number?\nstatistics calculator
Answer
Explanation:
Step1: Calculate the z - score
The z - score formula is $z=\frac{x-\mu}{\sigma}$, where $x = 5$, $\mu=7.5$, and $\sigma = 1.5$. $z=\frac{5 - 7.5}{1.5}=\frac{- 2.5}{1.5}\approx - 1.67$
Step2: Find the probability using the z - table
Looking up the z - score of $-1.67$ in the standard normal distribution table, the area to the left (probability $P(Z\lt - 1.67)$) is approximately $0.0475$. The probability of $Z\gt - 1.67$ is $P(Z\gt - 1.67)=1 - P(Z\lt - 1.67)=1 - 0.0475 = 0.9525$.
Step3: Calculate the expected number of oranges
We have a batch of $n = 2000$ oranges. The expected number of oranges that weigh more than 5 oz is $n\times P(Z\gt - 1.67)=2000\times0.9525 = 1905$.
Answer:
1905