A person is to install five devices each 6 7/8 inches wide with 3 1/2 inches between switches. how much space will be needed to
mount these switches?
1 answer:
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Answer:
(1) P( < 26 inches) = 0.0436
(2) P(27.5 inches < < 28.5 inches) = 0.2812
Step-by-step explanation:
We are given that the mean vertical leap of all NBA players is 28 inches. Suppose the standard deviation is 7 inches and 36 NBA players are selected at random.
Firstly, Let = mean vertical leap for the 36 players
Assuming the data follows normal distribution; so the z score probability distribution for sample mean is given by;
Z = ~ N(0,1)
where, = population mean vertical leap = 28 inches
= standard deviation = 7 inches
n = sample of NBA player = 36
(1) Probability that the mean vertical leap for the 36 players will be less than 26 inches is given by = P( < 26 inches)
P( < 26) = P(
<
) = P(Z < -1.71) = 1 - P(Z
1.71)
= 1 - 0.95637 = 0.0436
(2) <em>Now, here sample of NBA players is 26 so n = 26.</em>
Probability that the mean vertical leap for the 26 players will be between 27.5 and 28.5 inches is given by = P(27.5 inches < < 28.5 inches) = P(
< 28.5 inches) - P(
27.5 inches)
P( < 28.5) = P(
<
) = P(Z < 0.36) = 0.64058 {using z table}
P(
27.5) = P(
) = P(Z
-0.36) = 1 - P(Z < 0.36)
= 1 - 0.64058 = 0.35942
Therefore, P(27.5 inches < < 28.5 inches) = 0.64058 - 0.35942 = 0.2812
Answer:
Step-by-step explanation:
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