Question

the length of zebra pregnancies is normally distributed, with mean \\( \mu = 380 \\) and standard deviation \\( \sigma = 10 \\). find \\( p(x<375) \\).

enter your answer as an area under the curve with 4 decimal places.

\\( p(x<375)= \\)

(this problem is just like ch 6.1 so you are calculating a regular z - score and looking up the la.)

Explanation:

Step1: Calculate the z - score

The formula for the z - score is \(z=\frac{x-\mu}{\sigma}\).
Given \(\mu = 380\), \(\sigma=10\), and \(x = 375\).
Substitute the values into the formula: \(z=\frac{375 - 380}{10}=\frac{-5}{10}=- 0.5\)

Step2: Find the probability using the standard normal distribution table

We want to find \(P(X\lt375)\), which is equivalent to \(P(Z\lt - 0.5)\) in the standard normal distribution (\(Z\sim N(0,1)\)).
Looking up the value of \(z=-0.5\) in the standard normal distribution table (the area to the left of \(z\)), we get \(P(Z\lt - 0.5)=0.3085\)

Answer:

\(0.3085\)