Weights of women are normally distributed with mean 143 pounds and standard deviation 29 pounds. If 36 women are randomly selected, what is the probability that their mean weight is between 120 and 160 pounds? Round your answer to two decimal places.
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To find the probability that the mean weight of 36 randomly selected women is between 120 and 160 pounds, we can use the Central Limit Theorem.
According to the Central Limit Theorem, when a sample size is large enough (in this case, 36), the distribution of the sample mean will be approximately normally distributed, even if the population distribution is not normal.
First, we need to calculate the standard error, which is the standard deviation of the sampling distribution of the mean:
Standard Error = Standard Deviation / √(Sample Size)
Standard Error = 29 / √36
Standard Error = 29 / 6
Standard Error ≈ 4.83
Next, we can standardize the values of 120 and 160 using the formula:
Z = (X - μ) / σ,
where X is the value, μ is the mean, and σ is the standard deviation.
For 120 pounds:
Z1 = (120 - 143) / 4.83
Z1 ≈ -4.76
For 160 pounds:
Z2 = (160 - 143) / 4.83
Z2 ≈ 3.52
Now, we can find the probability using the standard normal distribution table.
P(120 < X < 160) = P(Z1 < Z < Z2)
P(-4.76 < Z < 3.52)
Looking up these values in the standard normal distribution table, we find:
P(Z < -4.76) ≈ 0 (very close to 0)
P(Z < 3.52) ≈ 0.9998
Therefore, P(-4.76 < Z < 3.52) ≈ 0.9998 - 0 ≈ 0.9998
So, the probability that the mean weight of 36 randomly selected women is between 120 and 160 pounds is approximately 0.9998 (or 99.98%) when rounded to two decimal places.
To solve this problem, we need to use the concept of the sampling distribution of the sample mean.
The sampling distribution of the sample mean is a probability distribution that describes the likelihood of obtaining different values of the sample mean from all possible samples taken from a population. The mean of the sampling distribution is equal to the population mean, and the standard deviation is equal to the population standard deviation divided by the square root of the sample size.
In this case, we know that the population mean (μ) is 143 pounds and the population standard deviation (σ) is 29 pounds. We are interested in finding the probability that the mean weight of a sample of 36 women (x̄) is between 120 and 160 pounds.
First, we need to calculate the standard deviation of the sampling distribution (σx̄). Since we have the population standard deviation (σ) and the sample size (n = 36), we can use the formula:
σx̄ = σ / √n
σx̄ = 29 / √36
= 29 / 6
≈ 4.83
Next, we need to convert the values of 120 and 160 pounds to z-scores using the formula:
z = (x - μ) / σ
For 120 pounds:
z1 = (120 - 143) / 4.83
≈ -4.76
For 160 pounds:
z2 = (160 - 143) / 4.83
≈ 3.52
Now, we can find the probabilities associated with these z-scores using a standard normal distribution table or a calculator.
Using the standard normal distribution table, the area to the left of z1 = -4.76 is close to 0, and the area to the left of z2 = 3.52 is approximately 0.9995. To find the probability between these two z-scores, we subtract the smaller area from the larger area:
P(-4.76 < z < 3.52) = 0.9995 - 0
= 0.9995
Therefore, the probability that the mean weight of a sample of 36 women is between 120 and 160 pounds is approximately 0.9995.