here are the summary statistics for the all - time top 50 womens long jump distances (in meters).\n|n|mean|sd…

1. Calculate the z - score for American Jackie Joyner - Kersee who jumped 7.49 meters in 1994

Explanation:

Step1: Recall z - score formula

The z - score formula is $z=\frac{x-\mu}{\sigma}$, where $x$ is the data point, $\mu$ is the mean, and $\sigma$ is the standard deviation. Given $\mu = 7.2$, $\sigma=0.12$, and $x = 7.49$.

Step2: Substitute values into formula

$z=\frac{7.49 - 7.2}{0.12}=\frac{0.29}{0.12}\approx2.42$

Answer:

$z\approx2.42$

2. How far did Jackie Joyner - Kersee jump in feet (1 meter = 3.28 feet)?

Explanation:

Step1: Use conversion factor

Multiply the distance in meters by the conversion factor. $d = 7.49\times3.28$

Step2: Calculate the result

$d=7.49\times3.28 = 24.5672$ feet

Answer:

$24.5672$ feet

3. If all 50 jumps were converted to feet, what would be the new mean and new standard deviation of the distances?

Explanation:

Step1: Convert the mean

Since we multiply each data - point by 3.28 to convert from meters to feet, the new mean $\mu_{new}=7.2\times3.28 = 23.616$ feet.

Step2: Convert the standard deviation

The standard deviation also gets multiplied by the conversion factor. So $\sigma_{new}=0.12\times3.28 = 0.3936$ feet.

Answer:

New mean: $23.616$ feet, New standard deviation: $0.3936$ feet

4. Calculate the z - score for Jackie Joyner - Kersee's jump when it is measured in feet. What do you notice?

Explanation:

Step1: Recall the new mean and standard deviation in feet

We found $\mu_{new}=23.616$ feet and $\sigma_{new}=0.3936$ feet, and $x_{new}=24.5672$ feet.

Step2: Calculate the z - score

$z=\frac{24.5672 - 23.616}{0.3936}=\frac{0.9512}{0.3936}\approx2.42$ We notice that the z - score is the same as when the measurement was in meters. This is because the z - score is a unit - less measure that standardizes the data point relative to the mean and standard deviation of the distribution, and multiplying all data points by a constant does not change the relative position of a particular data point within the distribution.

Answer:

$z\approx2.42$, and the z - score is the same as when measured in meters.

5. What is the variance for the distribution of dozens of eggs sold?

Explanation:

Step1: Recall the relationship between variance and standard deviation

The variance $\text{Var}(X)=\sigma^{2}$, where $\sigma$ is the standard deviation. Given $\sigma = 3.0$ dozen.

Step2: Calculate the variance

$\text{Var}(X)=3.0^{2}=9.0$ (dozen$^{2}$)

Answer:

$9.0$ (dozen$^{2}$)

6. Suppose the farmer converts all the values from dozens of eggs to number of eggs. What is the new mean and new standard deviation?

Explanation:

Step1: Recall the conversion factor

Since 1 dozen = 12 eggs, to convert from dozens to number of eggs, we multiply each data - point by 12. The original mean is $\mu = 14.7$ dozen. The new mean $\mu_{new}=14.7\times12=176.4$ eggs. The original standard deviation is $\sigma = 3.0$ dozen. The new standard deviation $\sigma_{new}=3.0\times12 = 36$ eggs.

Answer:

New mean: $176.4$ eggs, New standard deviation: $36$ eggs

7. What is the variance for the distribution of number of eggs sold?

Explanation:

Step1: Recall the new standard deviation in eggs

The new standard deviation $\sigma_{new}=36$ eggs.

Step2: Calculate the variance

Since $\text{Var}(X)=\sigma^{2}$, the new variance $\text{Var}_{new}=36^{2}=1296$ (eggs$^{2}$)

Answer:

$1296$ (eggs$^{2}$)

8. How does this new variance compare to the original variance?

Explanation:

Step1: Recall the original and new variances

The original variance was 9.0 (dozen$^{2}$). The new variance is 1296 (eggs$^{2}$). Since we multiplied each data - point by 12 to convert from dozens to eggs, and variance is the square of the standard deviation, the new variance is $12^{2}$ times the original variance. $9\times12^{2}=9\times144 = 1296$.

Answer:

The new variance is $144$ times the original variance.