Math

QuestionEstimate P(P( fewer than 3)) using normal approximation when np5np \geq 5 and nq5nq \geq 5 with n=14n=14 and p=0.4p=0.4. If not, state that normal approximation is not suitable.

Studdy Solution

STEP 1

Assumptions1. The number of trials, nn, is14. . The probability of success, pp, is0.4.
3. The probability of failure, qq, is1 - pp.
4. We want to find the probability of fewer than3 successes.
5. We are using the normal distribution as an approximation to the binomial distribution.

STEP 2

First, we need to check if the normal approximation is suitable. The normal approximation is suitable if np5np \geq5 and nq5nq \geq5.

STEP 3

Calculate npnp and nqnq.
np=n×p=14×0.np = n \times p =14 \times0.nq=n×q=14×(10.)nq = n \times q =14 \times (1 -0.)

STEP 4

Calculate the values of npnp and nqnq.
np=14×0.4=.6np =14 \times0.4 =.6nq=14×0.6=8.4nq =14 \times0.6 =8.4

STEP 5

Since both np5np \geq5 and nq5nq \geq5, we can use the normal approximation.

STEP 6

Next, we need to calculate the mean and standard deviation of the binomial distribution. The mean, μ\mu, is npnp and the standard deviation, σ\sigma, is npq\sqrt{npq}.

STEP 7

Calculate the mean and standard deviation.
μ=np=5.6\mu = np =5.6σ=npq=5.6×0.6\sigma = \sqrt{npq} = \sqrt{5.6 \times0.6}

STEP 8

Calculate the value of the standard deviation.
σ=5.6×0.6=1.8974\sigma = \sqrt{5.6 \times0.6} =1.8974

STEP 9

We want to find the probability of fewer than3 successes. In terms of the normal distribution, this is equivalent to finding (X<3)(X <3). However, we need to apply the continuity correction, so we actually need to find (X<2.5)(X <2.5).

STEP 10

To find this probability, we need to standardize the value2.5 using the formula Z=XμσZ = \frac{X - \mu}{\sigma}.

STEP 11

Calculate the standardized value.
Z=.55.6.8974Z = \frac{.5 -5.6}{.8974}

STEP 12

Calculate the value of ZZ.
Z=2.55.6.8974=.63Z = \frac{2.5 -5.6}{.8974} = -.63

STEP 13

Now we can find the probability (Z<.63)(Z < -.63) using the standard normal distribution table or a calculator.

STEP 14

Look up the value in the standard normal distribution table or calculate it using a calculator.
(Z<.63)=0.0516(Z < -.63) =0.0516So, the probability of fewer than3 successes is approximately0.0516.

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