a) Histogram (figure attached)
b) The expected number of defective is equal to 0.4 syringes.
c) Probability that the batch will be accepted is equal to 0.94275.
d) The standard deviation is 0.6164.
It is a binomial distribution "DISCRETE probability distribution that encapsulates the likelihood that a value will take one of two independent values when a particular set of conditions is met. The assumptions for the binomial distribution are that each trial has a single possible outcome, each trial has an equal chance of success, and each trial is independent of the others or mutually exclusive ".
Let X be the number of syringes being defective as a binomial random variable.
We are given that in the batch, there can be assumed to be 5% defective syringes.
∴ n = 8
Probability of syringes being defective, p = 0.05
Probability of syringes being not defective , q = 1-p = 1-0.05 = 0.95
a) Histogram (figure attached)
b) We know that the formula for Expectation in binomial distribution is;
μ =n×p = 8×0.05
=0.4 syringes
Therefore, the inspector will find the expected number of defective is equal to 0.4 syringes.
c) We have to find binomial probability for that we will use the binomial probability distribution is given by:
P(X=x) = ⁿCₓpˣ (1-p)^n-x, x = 0,1,2,3 ..8
Therefore,
P(x<2) = P(X=0) + P(X=1)
= ⁸C₀(0.05)⁰(1-0.05)⁸ + ⁸C₁(0.05)¹(1-0.05)⁷
= 0.66342 + 8×0.05×0.6983
= 0.66342 + 0.27933
= 0.94275
Therefore, probability that the batch will be accepted is equal to 0.94275.
d) We first find the variance:
σ² = Var(X) = n×p×(1-p)
= 8 × 0.05 ×(1-0.05)
= 0.38
Thus, the standard deviation is
σ = [tex]\sqrt{0.38}[/tex] = 0.6164
Thus,
a) Histogram (figure attached)
b) The expected number of defective is equal to 0.4 syringes.
c) Probability that the batch will be accepted is equal to 0.94275.
d) The standard deviation is 0.6164.
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