Question

20 POINTS

Answer 1

Let get all the information.
Mean:394
Deviation:35
Looking for At Least 380 feet
To find the z score you take the wanted probality subtract the mean and divide by the deviation.
z=(380-394)/35.
z=-0.4
So far so good.
But when he looked at the table he looked at probability that the home runs are less then 380 feet.
From this table
http://www.stat.ufl.edu/~athienit/Tables/Ztable.pdf
To find the answer you have to take 1-(the probability that the home runs are less than 380 feet)or 1-0.3446=0.6554
He made an error in his calculation.

Answer 2

Answer:

he made an error, the probability that his homeruns traveled at least 380 feet is 65.54%

Step-by-step explanation:

let´s see the Robinson's calculation:

using the formula for normal distribution:

Z = (x-μ)(σ/$\sqrt{n}$) [1]

where:

x is raw measurement

μ is the mean

σ is the standar deviation

n is the size of sample

in this case x = 380, μ = 394, σ = 35, n = 1.

then using the equation [1] he calculated the Z score:

$Z = \frac{380-394}{\frac{35}{\sqrt{1}}}\\\\Z = -\frac{14}{35}\\\\Z = -0.4$

and it's correct but his answer is wrong because he confuse the term at least.

at least means that the probability be greater than 380 feet, not less than 380 feet.

so he looked the Z score in a table of normal distribution (this table has been attached) and effectively this value is 0.3446. but as I told you this probability is for x < 380.

therefore, the probability that his homeruns traveled at least 380 feet is:

$P(x > 380) = 1 - P(x < 380)\\\\P(x > 380) = 1 - 0.3446\\\\P(x > 380) = 0.6554 \ or \ 65.54 \%$

please see the table attached.