A fitness company surveyed the weights of its 90 costumers. Assume that the weights are normally distrubted with a mean of 162 lbs and a standard deviation of 3.5 lbs. If you select one of its costumers randomly, what is the probability that the costumer's weight is
a) more than 162 lbs?
b) more than 155 lbs?
c) less than 172.5 lbs?
d) between 158.5 lbs and 169 lbs?
Use
http://davidmlane.com/hyperstat/z_table.html
again.
To solve this problem, we will use the properties of the normal distribution. The first step is calculating the z-score. The z-score measures the number of standard deviations an individual value is from the mean. We can then use the z-score to find the corresponding probability using the standard normal distribution table or a calculator.
a) Finding the probability that a customer's weight is more than 162 lbs:
To solve this, we need to find the area to the right of 162 lbs. We calculate the z-score as follows:
z = (x - μ) / σ
where x is the value (162), μ is the mean (162), and σ is the standard deviation (3.5).
Plugging in the values, we get:
z = (162 - 162) / 3.5
z = 0
Using the standard normal distribution table or a calculator, we find that the probability for a z-score of 0 is 0.5000. However, since we want the probability of a weight greater than 162 lbs, we need to find the area to the right of 0, which is given by 1 - 0.5000 = 0.5000.
Therefore, the probability that a customer's weight is more than 162 lbs is 0.5000 (or 50%).
b) Finding the probability that a customer's weight is more than 155 lbs:
Following the same steps, we need to find the area to the right of 155 lbs. Calculate the z-score as follows:
z = (155 - 162) / 3.5
z = -2
Using the standard normal distribution table or a calculator, we find that the probability for a z-score of -2 is approximately 0.0228. Since we want the probability of a weight greater than 155 lbs, we need to find the area to the right of -2, which is given by 1 - 0.0228 = 0.9772.
Therefore, the probability that a customer's weight is more than 155 lbs is approximately 0.9772 (or 97.72%).
c) Finding the probability that a customer's weight is less than 172.5 lbs:
For this case, we are interested in the area to the left of 172.5 lbs. Calculate the z-score:
z = (172.5 - 162) / 3.5
z = 3
Using the standard normal distribution table or a calculator, we find that the probability for a z-score of 3 is approximately 0.9987. Therefore, the probability that a customer's weight is less than 172.5 lbs is approximately 0.9987 (or 99.87%).
d) Finding the probability that a customer's weight is between 158.5 lbs and 169 lbs:
Here, we need to find the area between the two values. We calculate the z-scores for both ends:
For 158.5 lbs:
z1 = (158.5 - 162) / 3.5
z1 = -1
For 169 lbs:
z2 = (169 - 162) / 3.5
z2 = 2
Using the standard normal distribution table or a calculator, we find the probabilities:
For a z-score of -1, the probability is approximately 0.1587.
For a z-score of 2, the probability is approximately 0.9772.
Therefore, to find the probability between these two values, we subtract the smaller probability from the larger one: 0.9772 - 0.1587 = 0.8185.
Hence, the probability that a customer's weight is between 158.5 lbs and 169 lbs is approximately 0.8185 (or 81.85%).