contestada

Problem 3 Suppose that the distance of fly balls hit to the outfield (in baseball) is normally distributed with a mean of 300 feet and a standard deviation of 50 feet. Let X= distance in feet for a fly ball. In each appropriate box you are to enter either a rational number in "p/q" format or a decimal value accurate to the nearest 0.01 . X∼ ( 300 , 50 ) . For a random fly ball, what is the probability that this ball traveled fewer than 220 feet? P(X<220)= 0.05 . The 80th percentile of the distribution of fly balls is given by P(X< 342 )=0.80 .

Respuesta :

Answer:

P(X < 220) = 0.05

The 80th percentile of the distribution of fly balls is 342 feet.

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 question, we have that:

[tex]\mu = 300, \sigma = 50[/tex]

P(X<220)

This is the pvalue of Z when X = 220. So

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

[tex]Z = \frac{220 - 300}{50}[/tex]

[tex]Z = -1.6[/tex]

[tex]Z = -1.6[/tex] has a pvalue of 0.0548

Rounding to two decimal places

P(X < 220) = 0.05

80th percentile:

X when Z has a pvalue of 0.8. So X when Z = 0.84. So

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

[tex]0.84 = \frac{X - 300}{50}[/tex]

[tex]X - 300 = 50*0.84[/tex]

[tex]X = 342[/tex]

The 80th percentile of the distribution of fly balls is 342 feet.