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.
<em><u>Let X = actual capacity of the tanks</u></em>
SO, X ~ Normal()
The z-score probability distribution for normal distribution is given by;
Z = ~ N(0,1)
where, = mean capacity = 8544 gallons
= 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( < ) = P(Z < 0.50) = <u>0.6915</u>
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.