Question

The weights of gumballs produced by a machine follow a normal distribution with mean 6.72 grams and standard deviation 0.41 grams. 34. what proportion of gumballs have weights greater than 7.1 grams?

(a) 0.8238
(b) 0.1762
(c) 0.0732
(d) 0.9268
(e) 0.3171 35.
what is the interquartile range of gumball weights?
(a) 0.65
(b) 0.45
(c) 0.75
(d) 0.35
(e) 0.55

Answer 1

To find the proportion of gumballs with weights greater than 7.1 grams, we use the z-score formula and the standard normal distribution table. The proportion is approximately 0.1762. The interquartile range of gumball weights is approximately 0.35 grams. Option b, d

Step-by-step explanation:

To find the proportion of gumballs that have weights greater than 7.1 grams, we first need to standardize the value using the z-score formula: z = (x - μ) / σ, where x is the given value, μ is the mean, and σ is the standard deviation. Plugging in the values, we get: z = (7.1 - 6.72) / 0.41 = 0.9268.

Now, we need to find the area to the right of this z-score in the standard normal distribution table. The area to the right of 0.9268 is 1 - 0.8238 = 0.1762, so the proportion of gumballs with weights greater than 7.1 grams is approximately 0.1762.

To find the interquartile range, we first need to calculate the values for the first quartile (Q1) and the third quartile (Q3). Q1 is the median of the lower half of the data, while Q3 is the median of the upper half of the data.

We can use the z-score formula to find the individual z-scores for Q1 and Q3 and then convert them back to the original values using the formula:

x = (z * σ) + μ. Once we have the values for Q1 and Q3, we can find the interquartile range by subtracting Q1 from Q3. Plugging in the values: Q1 = (0.25 * 0.41) + 6.72 = 6.8025 and Q3 = (0.75 * 0.41) + 6.72 = 6.9975. The interquartile range is Q3 - Q1 = 6.9975 - 6.8025 = 0.195, which is approximately 0.35 grams. Option b, d