You survey 67 employees, and 30 respond that their job satisfaction is high. What percentage of your employees have high job sat
2 answers:
You might be interested in
Answer:
0.7486 = 74.86% observations would be less than 5.79
Step-by-step explanation:
I suppose there was a small typing mistake, so i am going to use the distribution as N (5.43,0.54)
Problems of normally distributed samples can be solved using the z-score formula.
In a set with mean and standard deviation
, the zscore of a measure X is given by:
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value 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. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
The general format of the normal distribution is:
N(mean, standard deviation)
Which means that:
What proportion of observations would be less than 5.79?
This is the pvalue of Z when X = 5.79. So
has a pvalue of 0.7486
0.7486 = 74.86% observations would be less than 5.79