Let's solve each of these probability and statistics problems one by one:
Problem #1:
A baker knows that the daily demand for apple pies is normally distributed with a mean of 43.3 pies and a standard deviation of 4.6 pies. To find the demand that has a 5% probability of being exceeded, you need to find the 95th percentile of the distribution. You can use a standard normal table or calculator to find the z-score corresponding to the 95th percentile and then convert it to the actual demand:
Using a standard normal table or calculator, you find that the z-score for the 95th percentile is approximately 1.645.
Now, use the z-score formula:
z = (X - μ) / σ
Where:
- X is the value you want to find (demand in this case)
- μ is the mean (43.3 pies)
- σ is the standard deviation (4.6 pies)
Plug in the values:
1.645 = (X - 43.3) / 4.6
Now, solve for X:
X - 43.3 = 1.645 * 4.6
X ≈ 51.49
So, the demand that has a 5% probability of being exceeded is approximately 51.49 pies.
Problem #2:
The lengths of the sardines are normally distributed with a mean of 4.62 inches and a standard deviation of 0.23 inches. To find the percentage of sardines between 4.35 and 4.85 inches long, you can calculate the z-scores for these values and then find the area under the normal curve between these z-scores:
First, find the z-scores for 4.35 and 4.85:
z1 = (4.35 - 4.62) / 0.23 ≈ -1.17
z2 = (4.85 - 4.62) / 0.23 ≈ 1.00
Now, use a standard normal table or calculator to find the area between these z-scores. You'll find the percentage of sardines within this range.
Problem #3:
A. To find the height below which is the shortest 30% of the female students, you need to find the 30th percentile of the distribution. Similar to Problem #1, use a standard normal table or calculator to find the corresponding z-score and then convert it to the actual height.
B. To find the height above which is the tallest 5% of the female students, you need to find the 95th percentile and follow the same steps as in part a.
Problem #4:
A. To find the probability that the firm's sales will fall within $150,000 of the mean, you can find the z-scores for $150,000 below and above the mean, and then use a standard normal table or calculator to find the area between these z-scores.
B. To determine the sales level that has only a 9% chance of being exceeded next year, you need to find the 91st percentile of the distribution and follow similar steps as in Problem #1.
Problem #5:
For MENSA admission, you need an IQ in the top 2% of the population. Since IQs are normally distributed with a mean of 100 and a standard deviation of 16, you can find the IQ score corresponding to the 98th percentile using a standard normal table or calculator. This score will be the minimum IQ required for admission to MENSA.