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An HP laser printer is advertised to print text documents at a speed of 18 ppm (pages per minute). The manufacturer tells you that the printing speed is actually a Normal random variable with a mean of 17.35 ppm and a standard deviation of 3.25 ppm. Suppose that you draw a random sample of 10 printers.

Required:
a. Using the information about the distribution of the printing speeds given by the manufacturer, find the probability that the mean printing speed of the sample is greater than 17.55 ppm.
b. Use normal approximation to find the probability that more than 48.6% of the sampled printers operate at the advertised speed (i.e. the printing speed is equal to or greater than 18 ppm)

User RRON
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1 Answer

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Answer:

a) The probability that the mean printing speed of the sample is greater than 17.55 ppm = 0.4247

b) The probability that more than 48.6% of the sampled printers operate at the advertised speed = 0.4197

Explanation:

The central limit theorem explains that for an independent random sample, the mean of the sampling distribution is approximately equal to the population mean and the standard deviation of the distribution of sample is given as

σₓ = (σ/√n)

where σ = population standard deviation

n = sample size

So,

Mean of the distribution of samples = population mean

μₓ = μ = 17.35 ppm

σₓ = (σ/√n) = (3.25/√10) = 1.028 ppm

a) The probability that the mean printing speed of the sample is greater than 17.55 ppm.

P(x > 17 55)

We first normalize 17.55 ppm

The standardized score for any value is the value minus the mean then divided by the standard deviation.

z = (x - μ)/σ = (17.55 - 17.35)/1.028 = 0.19

To determine the required probability

P(x > 17.55) = P(z > 0.19)

We'll use data from the normal probability table for these probabilities

P(x > 17.55) = P(z > 0.19) = 1 - P(z ≤ 0.19)

= 1 - 0.57535 = 0.42465 = 0.4247

b) The probability that more than 48.6% of the sampled printers operate at the advertised speed

We first find the probability that one randomly selected printer operates at the advertised speed.

Mean = 17.35 ppm

Standard deviation = 3.25 ppm

Advertised speed = 18 ppm

Required probability = P(x ≥ 18)

We standardize 18 ppm

z = (x - μ)/σ = (18 - 17.35)/3.25 = 0.20

To determine the required probability

P(x ≥ 18) = P(z ≥ 0.20)

We'll use data from the normal probability table for these probabilities

P(x ≥ 18) = P(z ≥ 0.20) = 1 - P(z < 0.20)

= 1 - 0.57926 = 0.42074

48.6% of the sample = 48.6% × 10 = 4.86

Greater than 4.86 printers out of 10 includes 5 upwards.

Probability that one printer operates at advertised speed = 0.42074

Probability that one printer does not operate at advertised speed = 1 - 0.42074 = 0.57926

probability that more than 48.6% of the sampled printers operate at the advertised speed will be obtained using binomial distribution formula since a binomial experiment is one in which the probability of success doesn't change with every run or number of trials. It usually consists of a number of runs/trials with only two possible outcomes, a success or a failure. The outcome of each trial/run of a binomial experiment is independent of one another.

Binomial distribution function is represented by

P(X = x) = ⁿCₓ pˣ qⁿ⁻ˣ

n = total number of sample spaces = 10

x = Number of successes required = greater than 4.86, that is, 5, 6, 7, 8, 9 and 10

p = probability of success = 0.42074

q = probability of failure = 0.57926

P(X > 4.86) = P(X ≥ 5) = P(X=5) + P(X=6) + P(X=7) + P(X=8) + P(X=9) + P(X=10) = 0.4196798909 = 0.4197

Hope this Helps!!!

User ATechGuy
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