Question

(1 point) Use the following information for all parts of this problem.The mean weight of an American newborn baby is 119.54 oz (about 7 pounds 8 ounces) with a population standarddeviation of 0.62 oz.In a random sample of 14 newborn babies, the mean weight was calculated to be 119.18 oz.

Answer 1

Given:

Mean, μ = 119.54 oz

Standard deviation, σ = 0.62 oz

Sample size, n = 14

Part 2.

Let's find the probability that the mean weight is at most 119.18 oz.

Here, we are to find:

P(X ≤ 119.18)

Apply the formula:

`P(X\leq\bar{x})=P(z\leq\frac{\bar{x}-\mu}{E})`

Where:

x' = 119.18

μ = 119.54

E = 0.1657

Thus, we have:

`\begin{gathered} P(X\leq119.18)=P(z\leq(119.18-119.54)/(0.1657)) \\ \\ P(X\leq119.18)=P(z\leq(-0.36)/(0.1657)) \\ \\ P(X\leq119.18)=P(z\leq−2.1726) \end{gathered}`

Using the Standard Normal Distribution Table, we have:

NORMSDIST(−2.1726) = 0.0149

Therefore, the probability that the mean weight is at most 119.18 is 0.0149

• ANSWER:

0.0149