Mathematics, 28.01.2020 20:51 yootmytoot

The diameter of a brand of tennis balls is approximately normally distributed, with a mean of 2.63 inches and a standard deviation of 0.03 inch. if you select a random sample of 9 tennis balls,

a. what is the sampling distribution of the mean?
b. what is the probability that the sample mean is less than 2.61 inches?
c. what is the probability that the sample mean is between 2.62 and 2.64 inches?
d. the probability is 6-% that the sample mean will be between what two values symmetrically distributed around the population mea

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Answer from: JKJKJKJKGFGH7470

a) The sampling distribution of the mean is 2.63 inches and sample standard deviation = 0.01

b) P(X < 2.61) = 0.0228

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

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

Let's solve this step by step:

Step 1: The diameter of a brand of tennis balls is approximately normally distributed.

Given:

Mean = 2.63 inches

Standard deviation = 0.03 inches

For random sample of 9 tennis balls

Step 2: Sample mean = 2.63 inches

Sample standard deviation = $\frac{\text{standard deviation}}{\sqrt{n} } =\frac{0.03}{\sqrt{9} } =\frac{0.03}{3} =0.01$

Step 3: The sample mean is less than 2.61 inches = P(X < 2.61) = 0.0228

Step 4: The probability that the sample mean is between 2.62 and 2.64 inches = P(2.62 < X < 2.64) = 0.6827

Step 5: 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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Answer from: maddiemalmstrom

a) sample mean = 2.63 inches

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

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

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)

Answer from: Quest

-2 is the right

Algebra semester 1 unit4 pretest linear equations

Answer from: Quest

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step-by-step explanation:

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