Question

A regulation baseball can weigh no more than 149 grams. A factory produces baseballs with weights that are normally distributed with a mean of 146 grams and a standard deviation of 2.3 grams. (a) If a baseball produced by the factory is randomly selected, what is the probability that it is within regulation weight?

Answer 1

Answer:

P(X$\leq$149) = 90.9%

Step-by-step explanation:

This is a normal distribution question. We need to find probability that randomly selected ball is within the regulation limit.

So,  

regulation is  =  X $\leq$ 149. Not greater than 149 grams.

Probability to find will be = P (X $\leq$ 149 )

For this we have the following formula.

z = $\frac{x - mean }{SD}$

SD = Standard Deviation = 2.3 grams

mean = 146 grams

x = 149

So,  

z = $\frac{149 - 146}{2.3}$

z = 1.304

Now, we have to use z-table to know what is the answer of P(X$\leq$149)

From the z - table if you see along 1.3 row and .04 column in the table. You will find 0.909 which is the probability that random selection of baseball will be within the regulation weight.

P(Z$\leq$149) = P(X$\leq$149) = 0.909 = 90.9%

Answer 2

Answer:

90.32% probability that it is within regulation weight

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$, the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

$\mu = 146, \sigma = 2.3$

If a baseball produced by the factory is randomly selected, what is the probability that it is within regulation weight?

Probability it weighs 149 or less grams, which is the pvalue of Z when X = 149. So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{149 - 146}{2.3}$

$Z = 1.30$

$Z = 1.30$ has a pvalue of 0.9032

90.32% probability that it is within regulation weight