Question

Assume that round widgets from a manufacturing process have diameters that are normally distributed with mean 135.75 centimeters and standard deviation 1.85 centimeters. The chances that a randomly selected widget has diameter between 129 centimeters and 143 centimeters is closest to which of the following?a. 99,999 out of 100,000.

b. 99 out of 100.
c. 999 out of 1,000.
d. 9,999 out of 10,000.

Answer 1

The probability that a randomly selected widget has a diameter between 129 cm and 143 cm is closest to 99,999 out of 100,000 based on the normal distribution with the given mean and standard deviation.

Step-by-step explanation:

To determine the probability that a randomly selected widget has a diameter between 129 centimeters and 143 centimeters given that the diameters of widgets are normally distributed with a mean of 135.75 centimeters and a standard deviation of 1.85 centimeters, we will apply the properties of the normal distribution.

First, we calculate the z-scores for 129 cm and 143 cm:
Z = (X - μ) / σ
Z1 = (129 - 135.75) / 1.85 = -3.65
Z2 = (143 - 135.75) / 1.85 = 3.92

Looking at a standard normal distribution table or using a calculator, we find that the area under the curve between these z-scores is very close to 1, which suggests the probability is very high. Given the options provided, the closest to the calculated probability would be almost 100%, which would most closely match:

a. 99,999 out of 100,000.

The other options significantly underestimate the probability for such a wide range within several standard deviations of the mean in a normal distribution.

Answer 2

b. 99 out of 100.

Step-by-step explanation:

We solve for this using z score formula

z-score formula is given as:

z = (x - μ)/σ

where

x is the raw score

μ is the population mean

σ is the population standard deviation.

a) x = 129cm, μ = 135.75cm, σ = 1.85cm

z = (129- 135.75)/1.85

z = -3.64865

Finding the Probability value of the z score from the z table:

P(x = 129) = P(z = -3.64865)

= 0.00013181

b) x = 143 cm, μ = 135.75cm, σ = 1.85cm

z = (143- 135.75)/1.85

z = 3.91892

Finding the Probability value of the z score from the z table:

P(x = 143) = P(z = 3.91892)

= 0.99996

The chances that a randomly selected widget has diameter between 129 centimeters and 143centimeters

= P(x = 143) - P(x = 129)

= P(z = 3.91892) - P(z = -3.64865)

= 0.99996 - 0.00013181

= 0.99982819

Converting to percentage

= 0.99982819 × 100

= 99.982819%

Therefore, option b. 99 out of 100 is the correct answer