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The probability that the sample average amount of time taken on each day is at most 11 min is 0.210.
Given data,
The time it takes a randomly chosen mortgage applicant to complete a certain form has a normal distribution with a mean value of 8 minutes and a standard deviation of 4 minutes. What is the likelihood that the sample average time taken on each day is at most 11 minutes if five people complete out a form on one day and six on another?
The normal distribution is a continuous distribution of data that has a bell-shaped curve. The normally distributed random variable x has to mean μ and standard deviation σ.
Also, the standard normal distribution represents a normal curve with a mean of 0 and a standard deviation of 1. Thus, the parameters involved in a normal distribution are mean μ and standard deviation σ.
Standardized z-score:
The standardized z-score represents the number of standard deviations the data point is away from the mean.
→ If the z-score takes a positive value when it is above the mean (0).
→ If the z-score takes a negative value when it is below the mean (0).
Let,
X ~ N = [x,σ/√n]
z = X - μ/ σ/√n
The probability that the sample average amount of time taken on each day is at most 11 min is obtained below,
From the given information, the sample size for the first day is 5 and the sample size for the second day is 6.
P(X ≤ 11 and Y ≤ 11) = P(X ≤ 11) * P(Y ≤ 11)
= P(z ≤ 1.67) * P(z ≤ 1.83)
= 0.452 * 0.466
= 0.210
Hence, the probability that the sample average amount of time taken on each day is at most 11 min is 0.210.
To learn more about probability click here:
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