A videotape store has an average weekly gross of $1,158 with a standard deviation of $120. Let x be the store's gross during a randomly selected week. If this is a normally distributed random variable, then the number of standard deviations from $1,158 to $1,360 is:

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Answer:

The number of standard deviations from $1,158 to $1,360 is 1.68.

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

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.

In this problem, we have that:

[tex]\mu = 1158, \sigma = 120[/tex]

The number of standard deviations from $1,158 to $1,360 is:

This is Z when X = 1360. So

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]Z = \frac{1360 - 1158}{120}[/tex]

[tex]Z = 1.68[/tex]

The number of standard deviations from $1,158 to $1,360 is 1.68.