The diameter of a brand of tennis balls is approximately normally​ distributed, with a mean of 2.56 inches and a standard deviation of 0.04

inch. A random sample of 11

tennis balls is selected. Complete parts​ (a) through​ (d) below.

sample mean = 2.63 inches

sample standard deviation = \frac{standard \hspace{0.15cm} deviation}{\sqrt{n} } = \frac{0.03}{\sqrt{9} } = \frac{0.03}{3} = 0.01

n

standarddeviation

=

9

0.03

=

3

0.03

=0.01

b) P(X < 2.61) = 0.0228

c.) P(2.62 < X < 2.64) = 0.6827

d.) Therefore 0.06 = P(2.6292 < X < 2.6307)

Step-by-step explanation:

i) the diameter of a brand of tennis balls is approximately normally distributed.

ii) mean = 2.63 inches

iii) standard deviation = 0.03 inches

iv) random sample of 9 tennis balls

v) sample mean = 2.63 inches

vi) sample standard deviation = \frac{standard \hspace{0.15cm} deviation}{\sqrt{n} } = \frac{0.03}{\sqrt{9} } = \frac{0.03}{3} = 0.01

n

standarddeviation

=

9

0.03

=

3

0.03

=0.01

vii) the sample mean is less than 2.61 inches = P(X < 2.61) = 0.0228

viii)the probability that the sample mean is between 2.62 and 2.64 inches

P(2.62 < X < 2.64) = 0.6827

ix) The probability is 6-% that the sample mean will be between what two values symmetrically distributed around the population measure

Therefore 0.06 = P(2.6292 < X < 2.6307)

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