Respuesta :
Answer:
0.7734; no, it does not.
Step-by-step explanation:
To find this probability we will use a z-score. We are dealing with the probability of a sample mean being larger than a given value; this means we use the formula
[tex]z=\frac{\bar{X}-\mu}{\sigma \div \sqrt{n}}[/tex]
Our x-bar in this problem is 155, as that is the sample mean we are trying to find the probability of. Our mean, μ, is 161 and our standard deviation, σ, is 31. Our sample size, n, is 15. This gives us
[tex]z=\frac{155-161}{31\div \sqrt{15}}=\frac{-6}{8.0042}=-0.75[/tex]
Using a z table, we see that the area under the curve less than this, corresponding with probability less than this value, is 0.2266. This means that the probability of the sample mean being larger than this is
1-0.2266 = 0.7734, or 77.34%.
Since there is a 77.34% chance of the passengers having a mean weight that is too heavy, this elevator is not safe.
The probability that the elevator is overloaded using the normal distribution principle is 0.7870
Given the Parameters :
- Mean, μ = 161
- Standard deviation, σ = 31
- Sample size, n = 15
- X = 155
Using the relation :
Zscore = (xbar - μ) ÷ (σ/√(n))
Zscore = (155 - 161) ÷ (31/√15)
Zscore = - 0.796
The probability that lift is overloaded can be expressed thus :
P(Z ≥ - 0.796) = 1 - P(Z ≤ - 0.796)
Using a normal distribution table ; PP(Z ≤ - 0.796) = 0.2130
P(Z ≥ - 0.796) = 1 - 0.2130
P(Z ≥ - 0.796) = 0.7870
Therefore, the probability that weight is overloaded is 0.7870
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