Question

A manufacturer makes golf balls whose weights average 1.62 ounces, with a standard deviation of 0.05 ounces. Estimate the probability that the weight of a group of 100 balls will lie in the interval 162 \pm 0.5 ounces.

Answer 1

Answer:

P(-1 < z < 1)  = 0.3174

Step-by-step explanation:

Mean (μ) = 1.62 ounces

Standard Deviation (σ) = 0.05

No of balls (sample size n) = 100

X = weight of a ball

Weight of a group of 100 balls must lie in the range 162 ± 0.5 ounces i.e. weight of a single ball will be 162/100 ± 0.5/100 ounces = 1.62 ± 0.005 ounces.

So, we need to find the probability P (1.615 < X < 1.625). We will use the central limit theorem.

z = (Χ' - μ)/(σ/$\sqrt{n}$)

P (1.615 < X < 1.625) = ($\frac{1.615 - 1.62}{0.05/\sqrt{100} }$ < (Χ - μ)/(σ/$\sqrt{n}$) < $\frac{1.625 - 1.62}{0.05/\sqrt{100} }$)

                                = (-1 < z < 1)

We need to find the probability of P (-1 < z < 1) by looking at the Normal Distribution Probability Table.

In order to make our working simpler, we need to break P (-1 < z < 1) into two parts: P(z < 1) and P(z > -1)

The probability for areas under the normal curve are given for P(z>X) so we can directly find the probability of P (z > -1) by referring to the normal probability table.

P(z > -1) = 0.1587

We can calculate P(z < 1) by subtracting P(z >1) from the total probability (i.e. 1). P(z >1) can be obtained from the normal probability table.

P(z < 1) = 1 - 0.8413 = 0.1587

By adding the two probabilities together, we get:

P(-1 < z < 1) = P(z < 1) + P (z > -1)

                   = 0.1587 + 0.1587

P(-1 < z < 1)  = 0.3174