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The height of sunflowers is Normally distributed with mean 50 inches and standard deviation 8 inches. Use your calculator's statistical functions to find the percent of all sunflowers that are between 35 and 40 inches tall. Be sure to write your answer as a percent, but don't put the percent symbol. Once in percent form, round to 2 places after the decimal.

Sagot :

Answer:

7.57

Step-by-step explanation:

When the distribution is normal, we use 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.

The height of sunflowers is Normally distributed with mean 50 inches and standard deviation 8 inches.

This means that [tex]\mu = 50, \sigma = 8[/tex]

Percent of all sunflowers that are between 35 and 40 inches tall.

As a proportion, this is the pvalue of Z when X = 40 subtracted by the pvalue of Z when X = 35. So

X = 40

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

[tex]Z = \frac{40 - 50}{8}[/tex]

[tex]Z = -1.25[/tex]

[tex]Z = -1.25[/tex] has a pvalue of 0.1057

X = 35

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

[tex]Z = \frac{35 - 50}{8}[/tex]

[tex]Z = -1.88[/tex]

[tex]Z = -1.88[/tex] has a pvalue of 0.03

0.1057 - 0.03 = 0.0757

As percent: 0.0757*100% = 7.57%