The distribution of annual earnings of all bank tellers with five years of experience is
skewed negatively. This distribution has a mean of Birr 15,000 and a standard deviation
of Birr 2000. If we draw a random sample of 30 tellers, what is the probability that their
earnings will average more than Birr 15,750 annually?
Required:
Calculate μ and
x
Calculate Z for X
Find the area covered by the interval
Interpret the results
please help me the answer
The distribution of annual earnings of all bank tellers with five years of experience is
skewed negatively. This distribution has a mean of Birr 15,000 and a standard deviation
of Birr 2000. If we draw a random sample of 30 tellers, what is the probability that their
earnings will average more than Birr 15,750 annually?
Required:
Calculate μ and
x
Calculate Z for X
Find the area covered by the interval
Interpret the results
0.7775
3.555
P = 1/60
The distribution of annual earnings of all bank tellers with five years of experience is skewed negatively. This distribution has a mean of Birr 15,000 and a standard deviation of Birr 2000. If we draw a random sample of 30 tellers, what is the probability that their earnings will average more than Birr 15,750 annually?
Required:
Calculate µ and x •
Calculate Z for X•
Please help me the answer as soon as possible
The distribution of annual earnings of all bank tellers with five years of experience is skewed negatively. This distribution has a mean of Birr 15,000 and a standard deviation of Birr 2000. If we draw a random sample of 30 tellers, what is the probability that their earnings will average more than Birr 15,750 annually.
• Calculate µ and ό×
• Calculate Z for X
• Find the area covered by the interval
• Interpret the results
Answers
To calculate the mean (µ) and standard deviation (σ), we use the following formulas:
µ = Mean
σ = Standard Deviation
n = Sample Size
µ = 15,000 Birr
σ = 2,000 Birr
n = 30 (Sample Size)
Next, we need to calculate the sample mean (x̄) and the standard deviation of the sample mean (σ x̄ ).
x̄ = Sample Mean
σ x̄ = Standard Deviation of Sample Mean
x̄ = µ = 15,000 Birr
σ x̄ = σ / sqrt(n)
Calculating σ x̄:
σ x̄ = 2,000 Birr / sqrt(30)
Once we have µ and σ x̄ , we can calculate the Z-score for X using the formula:
Z = (X - µ) / σ x̄
In this case, X is the desired earnings amount, which is 15,750 Birr.
Calculating Z-score:
Z = (15,750 - 15,000) / [2,000 / sqrt(30)]
Once we have Z, we can find the probability by looking up the Z-score in the standard normal distribution table or using statistical software.
Interpretation of the results:
The probability that the earnings of a random sample of 30 bank tellers with five years of experience will average more than 15,750 Birr annually can be obtained from the Z-score and the standard normal distribution table. It represents the likelihood of getting a sample mean of earnings greater than 15,750 Birr under the given conditions of the skewed distribution.