A manufacturer knows that their items have a lengths that are approximately normally distributed, with a mean of 9.8 inches, and
standard deviation of 1.6 inches. If 48 items are chosen at random, what is the probability that their mean length is greater than 9.2 inches
1 answer:
Answer:
0.6462
Step-by-step explanation:
Given that
Mean (m) = 9.8
Standard deviation (s) = 1.6
Probability that mean length is greater than 9.2
Z = (x - m) / s
Z = (9.2 - 9.8) / 1.6
Z = - 0.6 / 1.6
Zscore = - 0.375 = - 0.375
P(Z >- 0.375) = 1 - P(Z < - 0.375) ;
P(Z < - 0.375) = 0.3538
1 - P(Z < - 0.38) = 1 - 0.3538 = 0.6462
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Step-by-step explanation:
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s = ut + 0.5at^2
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