QUESTION IMAGE
Question
- the weights of elephants at birth can be modeled by a normal distribution with mean \\( \mu=200 \\) pounds and deviation \\( \sigma=25 \\) pounds. (a) label the horizontal axis using z-score values, \\( \mu=200 \\) pounds, and standard deviation \\( \sigma=25 \\) pounds. (b) calculate the z-score for the baby elephant weighing 230 pounds. (c) what percent of baby elephants weigh less than 230 pounds at birth? use the normal distribution z-score area table. answer: (d) a baby elephant, cardinal, was born in the cleveland zoo. only \\( 5 \\% \\) of elephants weigh more than cardinal did at birth. what will be z-score for cardinals weight? use the normal distribution z-score area table. (e) use cardinals z-score from (d) response. how much did this baby elephant cardinal weigh at birth?
Step1: Calculate the z - score for 230 pounds
The formula for the z - score is \(z=\frac{x-\mu}{\sigma}\), where \(x = 230\), \(\mu=200\), and \(\sigma = 25\).
Step2: Find the percentage of baby elephants weighing less than 230 pounds
Using the standard normal distribution table (z - score table), for \(z = 1.2\), the cumulative probability \(P(Z<1.2)\) is \(0.8849\) or \(88.49\%\)
Step3: Find the z - score for Cardinal
If only \(5\%\) of elephants weigh more than Cardinal, then \(P(Z>z)=0.05\), so \(P(Z\leq z)=1 - 0.05=0.95\). Looking up in the z - score table, the z - score corresponding to a cumulative probability of \(0.95\) is approximately \(z = 1.645\)
Step4: Calculate Cardinal's weight
Using the z - score formula \(z=\frac{x-\mu}{\sigma}\), and solving for \(x\). We know \(z = 1.645\), \(\mu = 200\), \(\sigma=25\)
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(b) The z - score is \(1.2\)
(c) Approximately \(88.49\%\)
(d) The z - score is approximately \(1.645\)
(e) Cardinal weighs approximately \(241.125\) pounds