# In a survey, 600 mothers and fathers were asked about the importance of sports for boys and girls. Of the parents interviewed, 70% said the genders are equal and should have equal opportunities to participate in sports.

B. Using the normal approximation without the continuity correction, sketch the probability distribution curve for the distribution of p-hat. Shade equal areas on both sides of the mean to show an area that represents a probability of .95, and label the upper and lower bounds of the shaded area as values of p-hat (not z-scores). Show your calculations for the upper and lower bounds. (3 points)

C. Considering the sketch in part B, the shaded area shows a .95 probability of what happening? In other words, what does the probability of .95 represent? (2 points)

D. Using the normal approximation, what's the probability a randomly drawn sample of parents of size 600 will have a sample proportion between 67% and 73%? Draw a sketch of the probability curve, shade the area representing the probability you're finding, and label the z-scores that represent the upper and lower bounds of the probability you're finding. Don't use the continuity correction. (4 points)

8 years ago

8 years ago

## 1. a. mean equals= .7 and standard deviation is 0.0187.

I don't know the rest...

1 year ago

## To answer these questions, we will use the concept of the normal distribution and the properties of sample proportions.

B. To sketch the probability distribution curve for the distribution of p-hat without the continuity correction, we need to calculate the upper and lower bounds of the shaded area that represents a probability of 0.95.

Since 70% of the parents interviewed said the genders are equal, the sample proportion for this group would be p-hat = 0.70. The population proportion is unknown, so we will use p-hat as an estimate.

To find the standard deviation of the sample proportion (sigma-p-hat), we can use the formula:

sigma-p-hat = sqrt((p-hat * (1 - p-hat)) / n),

where n is the sample size. In this case, n = 600.

sigma-p-hat = sqrt((0.70 * (1 - 0.70)) / 600)

= 0.0147 (rounded to four decimal places)

To find the z-scores corresponding to the upper and lower bounds of the shaded area, we can use the formula:

z = (p-hat - p) / sigma-p-hat,

where p is the population proportion, which we estimate with p-hat.

The z-score for the upper bound can be found using:

z-upper = (p-hat - p) / sigma-p-hat

= (0.70 - 0.50) / 0.0147

= 13.605 (rounded to three decimal places)

The z-score for the lower bound can be found using:

z-lower = (p-hat - p) / sigma-p-hat

= (0.70 - 0.50) / 0.0147

= 13.605 (rounded to three decimal places)

We can use a standard normal distribution table or a calculator to find the proportion associated with z-scores of 13.605. By looking up the z-scores in the table, we can find that the proportion is approximately 1 (or 100%). This means that the shaded area representing a probability of 0.95 would cover the entire distribution.

C. The probability of 0.95 represents the probability that the sample proportion falls within the shaded area, which in this case is the probability that the percentage of parents who say the genders are equal is between the upper and lower bounds we calculated.

D. To find the probability that a randomly drawn sample of parents of size 600 will have a sample proportion between 67% and 73%, we need to calculate the z-scores corresponding to these bounds.

The z-score for the upper bound can be found using:

z-upper = (p-hat - p) / sigma-p-hat

= (0.73 - 0.50) / 0.0147

= 19.728 (rounded to three decimal places)

The z-score for the lower bound can be found using:

z-lower = (p-hat - p) / sigma-p-hat

= (0.67 - 0.50) / 0.0147

= 11.565 (rounded to three decimal places)

Again, using a standard normal distribution table or a calculator, we can find the proportions associated with these z-scores. By looking up the z-scores in the table, we can find that the proportion for the upper bound (z-upper) is approximately 1 (or 100%) and the proportion for the lower bound (z-lower) is approximately 1 (or 100%). This means that the probability of randomly drawing a sample of parents of size 600 with a sample proportion between 67% and 73% is approximately 1 (or 100%).

1 year ago

## To solve the question, we will use the normal approximation to the binomial distribution.

A. The formula to find the standard deviation (sigma) for the sample proportion is given by:

sigma = sqrt((p * (1 - p)) / n)

Where:

p = proportion of the population (0.70)

n = sample size (600)

Substituting the values in the formula, we get:

sigma = sqrt((0.70 * (1 - 0.70)) / 600)

sigma â‰ˆ 0.014

Now, to find the value of z for a probability of 0.95, we will use the z-table. The z-table provides the z-values corresponding to particular probabilities. For a two-tailed test, subtract the desired probability (0.95) from 1 and divide by 2 to get the tail probability (0.025).

Using the z-table, the z-value corresponding to a right-tail probability of 0.025 is approximately 1.96.

To find the upper and lower bounds of the shaded area, we use the following formulas:

upper bound = p + z * sigma

lower bound = p - z * sigma

Substituting the values, we get:

upper bound = 0.70 + 1.96 * 0.014 â‰ˆ 0.728

lower bound = 0.70 - 1.96 * 0.014 â‰ˆ 0.672

B. The shaded area represents the probability of getting a sample proportion (p-hat) between 0.672 and 0.728, given a sample size of 600. In other words, it represents the probability that, in a random sample of 600 parents, the proportion who believe the genders are equal and should have equal opportunities to participate in sports falls between 0.672 and 0.728.

C. The probability of 0.95 in part B represents the confidence level or the level of certainty that the sample proportion falls between the upper and lower bounds. In this case, we can say with 95% confidence that the true proportion of parents who believe the genders are equal and should have equal opportunities to participate in sports lies between 0.672 and 0.728.

D. To find the probability of a sample proportion between 67% and 73%, we use the same approach as in part B.

First, we find the z-values corresponding to the upper and lower bounds:

upper z = (0.73 - 0.70) / 0.014 â‰ˆ 2.14

lower z = (0.67 - 0.70) / 0.014 â‰ˆ -2.14

Using the z-table, we find that the area to the right of z = 2.14 is approximately 0.0166, and the area to the left of z = -2.14 is also approximately 0.0166.

To find the probability of a sample proportion between 67% and 73%, we calculate the area between these two z-values:

probability = 1 - (0.0166 + 0.0166) â‰ˆ 0.9668

Therefore, the probability that a randomly drawn sample of parents of size 600 will have a sample proportion between 67% and 73% is approximately 0.9668.