A manufacturing facility produces iron rods whose mean diameter is 2.5 inches with a standard deviation of 0.2 inch. If a sample of 50 metal rods produced by the facility were randomly selected, what is the probability that the mean of these rods will lie between 2.4 inches and 2.8 inches?
According to census statistics, Hispanics comprise 27.5% of the population of New York City. Suppose a random sample of 400 people is selected from New York City. What is the probability that there are fewer than 22% of Hispanics in the sample? Hint: Use Normal approximation to the Binomial distribution.
Use z-scores.
First problem:
z = (x - mean)/(sd/√n)
Find two z-scores:
z = (2.4 - 2.5)/(0.2/√50) = ?
z = (2.8 - 2.5)/(0.2/√50) = ?
Finish the calculations.
Next, check a z-table to find the probability between the two scores.
Find mean and standard deviation.
mean = np = 400 * .275 = ?
standard deviation = √npq = √(400 * .275 * .725) = ?
Note: q = 1 - p
Finish the calculations.
Next, use z-scores:
z = (x - mean)/sd
z = (.22 - mean)/sd
Substitute the mean and standard deviation into the z-score formula and calculate, then use a z-table to find the probability.
I hope this will help get you started.
Note: For the second problem, we use the normal approximation to the binomial distribution when finding the mean and standard deviation.
How would you check the Z table for the first problem?
To find the probability that the mean diameter of the metal rods lies between 2.4 inches and 2.8 inches, we can use the Central Limit Theorem (CLT) since the population standard deviation is known.
The Central Limit Theorem states that for a large enough sample size (typically considered to be 30 or more), the sampling distribution of the sample mean will be approximately normally distributed, regardless of the shape of the population distribution.
First, we need to calculate the standard error of the mean, which is the standard deviation of the sampling distribution. The formula for the standard error (SE) is:
SE = population standard deviation / sqrt(sample size)
In this case, the population standard deviation is 0.2 inch, and the sample size is 50. Therefore, the standard error is:
SE = 0.2 / sqrt(50) = 0.0283
Next, we need to standardize the values 2.4 and 2.8 using the formula for z-score:
z = (x - mean) / standard error
For 2.4 inches:
z1 = (2.4 - 2.5) / 0.0283 = -3.53
For 2.8 inches:
z2 = (2.8 - 2.5) / 0.0283 = 10.61
To find the probability that the mean diameter lies between 2.4 and 2.8 inches, we need to calculate the area under the normal distribution curve between these z-values. We can do this by looking up the z-values in a standard normal distribution table or by using a statistical software or calculator.
P(2.4 < mean < 2.8) = P(-3.53 < z < 10.61)
Now we need to find the probability associated with each z-value using the standard normal distribution table or a statistical software.
P(z < -3.53) = 0.00017 (approximately)
P(z < 10.61) = 1 (since it is beyond the right tail of the distribution, essentially 100%)
Finally, we calculate the probability of the mean diameter lying between 2.4 and 2.8 inches by subtracting the smaller probability from the larger one:
P(2.4 < mean < 2.8) = P(z < 10.61) - P(z < -3.53) ≈ 1 - 0.00017 ≈ 0.9998
Therefore, the probability that the mean diameter of the metal rods lies between 2.4 inches and 2.8 inches is approximately 0.9998, or 99.98%.
For the second question regarding the probability of fewer than 22% Hispanics in the sample, we can use the Normal approximation to the Binomial distribution. When the sample size is large and the success probability is not extremely small or large, the binomial distribution can be approximated by a normal distribution.
The mean of the binomial distribution is given by n * p, where n is the sample size and p is the success probability. In this case, n is 400 (sample size) and p is 0.275 (percentage of Hispanics in the population).
The standard deviation of the binomial distribution is given by sqrt(n * p * (1 - p)). Using the given values, the standard deviation is:
SD = sqrt(400 * 0.275 * (1 - 0.275)) ≈ 8.839
Now, to find the probability of fewer than 22% Hispanics, we need to standardize this value using the formula for z-score:
z = (x - mean) / standard deviation
With x = 0.22 (22%) and the mean given as 400 * 0.275, we can calculate the z-score:
z = (0.22 - 400 * 0.275) / 8.839 ≈ -1.793
Next, we need to find the probability associated with this z-value, which represents the area to the left of this value on the standard normal distribution.
P(z < -1.793) can be found by referencing a standard normal distribution table or using a statistical software. Let's assume it is approximately 0.036 (approximately).
Therefore, the probability that there are fewer than 22% Hispanics in the sample is approximately 0.036, or 3.6%.