Respuesta :
Answer:
(a) The z-score corresponding to a tank with a capacity of 8550 gallons is z = 0.50.
(b) The required probability is 0.6915.
Step-by-step explanation:
We are given that the actual capacity of the tanks is normally distributed with mean 8544 gallons and standard deviation 12 gallons.
The actual capacity of the tanks is normally distributed.
Let X = actual capacity of the tanks
SO, X ~ Normal([tex]\mu=8544,\sigma^{2} =12^{2}[/tex])
The z-score probability distribution for normal distribution is given by;
Z = [tex]\frac{X-\mu}{\sigma}[/tex] ~ N(0,1)
where, [tex]\mu[/tex] = mean capacity = 8544 gallons
[tex]\sigma[/tex] = standard deviation = 12 gallons
The Z-score measures how many standard deviations the measure is away from the mean. After finding the Z-score, we look at the z-score table and find the p-value (area) associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X.
Now, Probability that a randomly selected tank will have a capacity of less than 8550 gallons is given by = P(X < 8550 gallons)
P(X < 8550 gallons) = P( [tex]\frac{X-\mu}{\sigma}[/tex] < [tex]\frac{8550-8544}{12}[/tex] ) = P(Z < 0.50) = 0.6915
The above probability is calculated by looking at the value of x = 0.50 in the z table which has an area of 0.6915.
(a) The z-score corresponding to a tank with a capacity of 8550 gallons is z = 0.50.
(b) The required probability is 0.6915.
