Question

The time required to cook a pizza at a neighborhood pizza joint is normally distributed with a mean of 12 minutes and a standard deviation of 2 minutes. find the time for each event. (round your answers to 2 decimal places.)

a. highest 5 percent min.
b. lowest 50 percent min.
c. middle 95 percent min. to min.
d. lowest 80 percent min.

Answer 1

a. The highest 5 percent time is approximately 15.29 minutes. b. The lowest 50 percent time is 12 minutes. c. The middle 95 percent interval is approximately 8.08 minutes to 15.92 minutes. d. The lowest 80 percent time is approximately 10.32 minutes.

Step-by-step explanation:

a. To find the highest 5 percent, we need to find the value that separates the highest 5 percent from the rest. This value is called the upper 5th percentile. Using the z-score formula, z = (x - mean) / standard deviation, we can find the z-score associated with the upper 5th percentile. By looking up this z-score in the z-table, we can find the corresponding value. In this case, the z-score is approximately 1.645. Using the formula, x = z * standard deviation + mean, we can calculate the time needed for the highest 5 percent: x = 1.645 * 2 + 12 = 15.29 minutes.

b. To find the lowest 50 percent, we need to find the value that separates the lowest 50 percent from the rest. This value is called the median. The median of a normally distributed data set is equal to the mean. Hence, the lowest 50 percent time is the mean of 12 minutes.

c. To find the time for the middle 95 percent interval, we need to find the values that separate the middle 95 percent from the rest. These values are called the lower 2.5th percentile and the upper 2.5th percentile. Using the z-score formula, we can find the z-scores associated with the lower and upper 2.5th percentiles. By looking up these z-scores in the z-table, we can find the corresponding values. In this case, the z-scores are approximately -1.960 and 1.960. Using the formula, x = z * standard deviation + mean, we can calculate the times: lower 2.5th percentile = -1.960 * 2 + 12 = 8.08 minutes, and upper 2.5th percentile = 1.960 * 2 + 12 = 15.92 minutes.

d. To find the lowest 80 percent, we need to find the value that separates the lowest 80 percent from the rest. This value is called the lower 20th percentile. Using the z-score formula, we can find the z-score associated with the lower 20th percentile. By looking up this z-score in the z-table, we can find the corresponding value. In this case, the z-score is approximately -0.841. Using the formula, x = z * standard deviation + mean, we can calculate the time needed for the lowest 80 percent: x = -0.841 * 2 + 12 = 10.32 minutes.

Answer 2

Focus on "middle 95 percent min. to min" first. The central 95% of the area under the std. normal curve represents -2 std. dev. up to 2 std. dev. We need to convert std. dev. here into the actual x (minute) values, using the z-score formula:
x - 12
z = -------------- Let this equal -2 first and find x; then let this equal 2, and find x
2
x - 12
x again: -2 = ---------- => -4 = x - 12 => 8 = x
2

This means, when we're considering the middle 95% of cases, the minimum cooking time is 8 minutes. The max would then be 12+4 = 16 minutes.

Hope this helps you get started. Try solving the other 3 parts. Let me know if you need further help with this.